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7.6: Basic Concepts of Probability

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Learning Objectives

After completing this section, you should be able to:

• Define probability including impossible and certain events.
• Calculate basic theoretical probabilities.
• Calculate basic empirical probabilities.
• Distinguish among theoretical, empirical, and subjective probability.
• Calculate the probability of the complement of an event.

It all comes down to this. The game of Monopoly that started hours ago is in the home stretch. Your sister has the dice, and if she rolls a 4, 5, or 7 she’ll land on one of your best spaces and the game will be over. How likely is it that the game will end on the next turn? Is it more likely than not? How can we measure that likelihood? This section addresses this question by introducing a way to measure uncertainty.

Introducing Probability

Uncertainty is, almost by definition, a nebulous concept. In order to put enough constraints on it that we can mathematically study it, we will focus on uncertainty strictly in the context of experiments. Recall that experiments are processes whose outcomes are unknown; the sample space for the experiment is the collection of all those possible outcomes. When we want to talk about the likelihood of particular outcomes, we sometimes group outcomes together; for example, in the Monopoly example at the beginning of this section, we were interested in the roll of 2 dice that might fall as a 4, 5, or 7. A grouping of outcomes that we’re interested in is called an event . In other words, an event is a subset of the sample space of an experiment; it often consists of the outcomes of interest to the experimenter.

Once we have defined the event that interests us, we can try to assess the likelihood of that event. We do that by assigning a number to each event ( E E ) called the probability of that event ( P ( E ) P ( E ) ). The probability of an event is a number between 0 and 1 (inclusive). If the probability of an event is 0, then the event is impossible. On the other hand, an event with probability 1 is certain to occur. In general, the higher the probability of an event, the more likely it is that the event will occur.

Example 7.16

Determining certain and impossible events.

Consider an experiment that consists of rolling a single standard 6-sided die (with faces numbered 1-6). Decide if these probabilities are equal to zero, equal to one, or somewhere in between.

• P ( roll a 4 ) P ( roll a 4 )
• P ( roll a 7 ) P ( roll a 7 )
• P ( roll a positive number ) P ( roll a positive number )
• P ( roll a 1 3 ) P ( roll a 1 3 )
• P ( roll an even number ) P ( roll an even number )
• P ( roll a single-digit number ) P ( roll a single-digit number )

Let's start by identifying the sample space. For one roll of this die, the possible outcomes are {1, 2, 3, 4, 5,6}. We can use that to assess these probabilities:

• We see that 4 is in the sample space, so it’s possible that it will be the outcome. It’s not certain to be the outcome, though. So, 0 < P ( roll a 4 ) < 1 0 < P ( roll a 4 ) < 1 .
• Notice that 7 is not in the sample space. So, P ( roll a 7 ) = 0 P ( roll a 7 ) = 0 .
• Every outcome in the sample space is a positive number, so this event is certain. Thus, P ( roll a positive number ) = 1 P ( roll a positive number ) = 1 .
• Since 1 3 1 3 is not in the sample space, P ( roll a 1 3 ) = 0 P ( roll a 1 3 ) = 0 .
• Some outcomes in the sample space are even numbers (2, 4, and 6), but the others aren’t. So, 0 < P ( roll an even number ) < 1 0 < P ( roll an even number ) < 1 .
• Every outcome in the sample space is a single-digit number, so P ( roll a single-digit number ) = 1 P ( roll a single-digit number ) = 1 .

Your Turn 7.16

Three ways to assign probabilities.

The probabilities of events that are certain or impossible are easy to assign; they’re just 1 or 0, respectively. What do we do about those in-between cases, for events that might or might not occur? There are three methods to assign probabilities that we can choose from. We’ll discuss them here, in order of reliability.

Method 1: Theoretical Probability

The theoretical method gives the most reliable results, but it cannot always be used. If the sample space of an experiment consists of equally likely outcomes, then the theoretical probability of an event is defined to be the ratio of the number of outcomes in the event to the number of outcomes in the sample space.

For an experiment whose sample space S S consists of equally likely outcomes, the theoretical probability of the event E E is the ratio

P ( E ) = n ( E ) n ( S ) , P ( E ) = n ( E ) n ( S ) ,

where n ( E ) n ( E ) and n ( S ) n ( S ) denote the number of outcomes in the event and in the sample space, respectively.

Example 7.17

Computing theoretical probabilities.

Recall that a standard deck of cards consists of 52 unique cards which are labeled with a rank (the whole numbers from 2 to 10, plus J, Q, K, and A) and a suit ( ♣ ♣ , ♢ ♢ , ♡ ♡ , or ♠ ♠ ). A standard deck is thoroughly shuffled, and you draw one card at random (so every card has an equal chance of being drawn). Find the theoretical probability of each of these events:

• The card is 10 ♠ 10 ♠ .
• The card is a ♡ ♡ .
• The card is a king (K).

There are 52 cards in the deck, so the sample space for each of these experiments has 52 elements. That will be the denominator for each of our probabilities.

• There is only one 10 ♠ 10 ♠ in the deck, so this event only has one outcome in it. Thus, P ( 10 ♠ ) = 1 52 P ( 10 ♠ ) = 1 52 .
• There are 13 ♡ s ♡ s in the deck, so P ( ♡ ) = 13 52 = 1 4 P ( ♡ ) = 13 52 = 1 4 .
• There are 4 cards of each rank in the deck, so P ( K ) = 4 52 = 1 13 P ( K ) = 4 52 = 1 13 .

Your Turn 7.17

It is critical that you make sure that every outcome in a sample space is equally likely before you compute theoretical probabilities!

Example 7.18

Using tables to find theoretical probabilities.

In the Basic Concepts of Probability, we were considering a Monopoly game where, if your sister rolled a sum of 4, 5, or 7 with 2 standard dice, you would win the game. What is the probability of this event? Use tables to determine your answer.

We should think of this experiment as occurring in two stages: (1) one die roll, then (2) another die roll. Even though these two stages will usually occur simultaneously in practice, since they’re independent, it’s okay to treat them separately.

Step 1: Since we have two independent stages, let’s create a table (Figure 7.27), which is probably the most efficient method for determining the sample space.

Now, each of the 36 ordered pairs in the table represent an equally likely outcome.

Step 2: To make our analysis easier, let’s replace each ordered pair with the sum (Figure 7.28).

Step 3: Since the event we’re interested in is the one consisting of rolls of 4, 5, or 7. Let’s shade those in (Figure 7.29).

Our event contains 13 outcomes, so the probability that your sister rolls a losing number is 13 36 13 36 .

Your Turn 7.18

Example 7.19, using tree diagrams to compute theoretical probability.

If you flip a fair coin 3 times, what is the probability of each event? Use a tree diagram to determine your answer

• You flip exactly 2 heads.
• You flip 2 consecutive heads at some point in the 3 flips.
• All 3 flips show the same result.

Let’s build a tree to identify the sample space (Figure 7.30).

The sample space is {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, which has 8 elements.

• Flipping exactly 2 heads occurs three times (HHT, HTH, THH), so the probability is 3 8 3 8 .
• Flipping 2 consecutive heads at some point in the experiment happens 3 times: HHH, HHT, THH. So, the probability is 3 8 3 8 .
• There are 2 outcomes that all show the same result: HHH and TTT. So, the probability is 2 8 = 1 4 2 8 = 1 4 .

Your Turn 7.19

People in mathematics, gerolamo cardano.

The first known text that provided a systematic approach to probabilities was written in 1564 by Gerolamo Cardano (1501–1576). Cardano was a physician whose illegitimate birth closed many doors that would have otherwise been open to someone with a medical degree in 16th-century Italy. As a result, Cardano often turned to gambling to help ends meet. He was a remarkable mathematician, and he used his knowledge to gain an edge when playing at cards or dice. His 1564 work, titled Liber de ludo aleae (which translates as Book on Games of Chance ), summarized everything he knew about probability. Of course, if that book fell into the hands of those he played against, his advantage would disappear. That’s why he never allowed it to be published in his lifetime (it was eventually published in 1663). Cardano made other contributions to mathematics; he was the first person to publish the third degree analogue of the Quadratic Formula (though he didn’t discover it himself), and he popularized the use of negative numbers.

Method 2: Empirical Probability

Theoretical probabilities are precise, but they can’t be found in every situation. If the outcomes in the sample space are not equally likely, then we’re out of luck. Suppose you’re watching a baseball game, and your favorite player is about to step up to the plate. What is the probability that he will get a hit?

In this case, the sample space is {hit, not a hit}. That doesn’t mean that the probability of a hit is 1 2 1 2 , since those outcomes aren’t equally likely. The theoretical method simply can’t be used in this situation. Instead, we might look at the player’s statistics up to this point in the season, and see that he has 122 hits in 531 opportunities. So, we might think that the probability of a hit in the next plate appearance would be about 122 531 ≈ 0.23 122 531 ≈ 0.23 . When we use the outcomes of previous replications of an experiment to assign a probability to the next replication, we’re defining an empirical probability . Empirical probability is assigned using the outcomes of previous replications of an experiment by finding the ratio of the number of times in the previous replications the event occurred to the total number of previous replications.

Empirical probabilities aren’t exact, but when the number of previous replications is large, we expect them to be close. Also, if the previous runs of the experiment are not conducted under the exact set of circumstances as the one we’re interested in, the empirical probability is less reliable. For instance, in the case of our favorite baseball player, we might try to get a better estimate of the probability of a hit by looking only at his history against left- or right-handed pitchers (depending on the handedness of the pitcher he’s about to face).

Probability and Statistics

One of the broad uses of statistics is called statistical inference, where statisticians use collected data to make a guess (or inference) about the population the data were collected from. Nearly every tool that statisticians use for inference is based on probability. Not only is the method we just described for finding empirical probabilities one type of statistical inference, but some more advanced techniques in the field will give us an idea of how close that empirical probability might be to the actual probability!

Example 7.20

Finding empirical probabilities.

Assign an empirical probability to the following events:

• Jose is on the basketball court practicing his shots from the free throw line. He made 47 out of his last 80 attempts. What is the probability he makes his next shot?
• Amy is about to begin her morning commute. Over her last 60 commutes, she arrived at work 12 times in under half an hour. What is the probability that she arrives at work in 30 minutes or less?
• Felix is playing Yahtzee with his sister. Felix won 14 of the last 20 games he played against her. How likely is he to win this game?
• Since Jose made 47 out of his last 80 attempts, assign this event an empirical probability of 47 80 ≈ 59 % 47 80 ≈ 59 % .
• Amy completed the commute in under 30 minutes in 12 of the last 60 commutes, so we can estimate her probability of making it in under 30 minutes this time at 12 60 = 20 % 12 60 = 20 % .
• Since Felix has won 14 of the last 20 games, assign a probability for a win this time of 14 20 = 70 % 14 20 = 70 % .

Your Turn 7.20

Work it out, buffon’s needle.

A famous early question about probability (posed by Georges-Louis Leclerc, Comte de Buffon in the 18th century) had to do with the probability that a needle dropped on a floor finished with wooden slats would lay across one of the seams. If the distance between the slats is exactly the same length as the needle, then it can be shown using calculus that the probability that the needle crosses a seam is 2 π 2 π . Using toothpicks or matchsticks (or other uniformly long and narrow objects), assign an empirical probability to this experiment by drawing parallel lines on a large sheet of paper where the distance between the lines is equal to the length of your dropping object, then repeatedly dropping the objects and noting whether the object touches one of the lines. Once you have your empirical probability, take its reciprocal and multiply by 2. Is the result close to π π ?

Method 3: Subjective Probability

In cases where theoretical probability can’t be used and we don’t have prior experience to inform an empirical probability, we’re left with one option: using our instincts to guess at a subjective probability . A subjective probability is an assignment of a probability to an event using only one’s instincts.

Subjective probabilities are used in cases where an experiment can only be run once, or it hasn’t been run before. Because subjective probabilities may vary widely from person to person and they’re not based on any mathematical theory, we won’t give any examples. However, it’s important that we be able to identify a subjective probability when we see it; they will in general be far less accurate than empirical or theoretical probabilities.

Example 7.21

Distinguishing among theoretical, empirical, and subjective probabilities.

Classify each of the following probabilities as theoretical, empirical, or subjective.

• An eccentric billionaire is testing a brand new rocket system. He says there is a 15% chance of failure.
• With 4 seconds to go in a close basketball playoff game, the home team need 3 points to tie up the game and send it to overtime. A TV commentator says that team captain should take the final 3-point shot, because he has a 38% chance of making it (greater than every other player on the team).
• Felix is losing his Yahtzee game against his sister. He has one more chance to roll 2 dice; he’ll win the game if they both come up 4. The probability of this is about 2.8%.
• This experiment has never been run before, so the given probability is subjective.
• Presumably, the commentator has access to each player’s performance statistics over the entire season. So, the given probability is likely empirical.
• Rolling 2 dice results in a sample space with equally likely outcomes. This probability is theoretical. (We’ll learn how to calculate that probability later in this chapter.)

Your Turn 7.21

Benford’s law.

In 1938, Frank Benford published a paper (“The law of anomalous numbers,” in Proceedings of the American Philosophical Society ) with a surprising result about probabilities. If you have a list of numbers that spans at least a couple of orders of magnitude (meaning that if you divide the largest by the smallest, the result is at least 100), then the digits 1–9 are not equally likely to appear as the first digit of those numbers, as you might expect. Benford arrived at this conclusion using empirical probabilities; he found that 1 was about 6 times as likely to be the initial digit as 9 was!

New Probabilities from Old: Complements

One of the goals of the rest of this chapter is learning how to break down complicated probability calculations into easier probability calculations. We’ll look at the first of the tools we can use to accomplish this goal in this section; the rest will come later.

Given an event E E , the complement of E E (denoted E ′ E ′ ) is the collection of all of the outcomes that are not in E E . (This is language that is taken from set theory, which you can learn more about elsewhere in this text.) Since every outcome in the sample space either is or is not in E E , it follows that n ( E ) + n ( E ′ ) = n ( S ) n ( E ) + n ( E ′ ) = n ( S ) . So, if the outcomes in S S are equally likely, we can compute theoretical probabilities P ( E ) = n ( E ) n ( S ) P ( E ) = n ( E ) n ( S ) and P ( E ′ ) = n ( E ′ ) n ( S ) P ( E ′ ) = n ( E ′ ) n ( S ) . Then, adding these last two equations, we get

P ( E ) + P ( E ′ ) = n ( E ) n ( S ) + n ( E ′ ) n ( S ) = n ( E ) + n ( E ′ ) n ( S ) = n ( S ) n ( S ) = 1 P ( E ) + P ( E ′ ) = n ( E ) n ( S ) + n ( E ′ ) n ( S ) = n ( E ) + n ( E ′ ) n ( S ) = n ( S ) n ( S ) = 1

Thus, if we subtract P ( E ′ ) P ( E ′ ) from both sides, we can conclude that P ( E ) = 1 − P ( E ′ ) P ( E ) = 1 − P ( E ′ ) . Though we performed this calculation under the assumption that the outcomes in S S are all equally likely, the last equation is true in every situation.

P ( E ) = 1 − P ( E ′ ) P ( E ) = 1 − P ( E ′ )

How is this helpful? Sometimes it is easier to compute the probability that an event won’t happen than it is to compute the probability that it will . To apply this principle, it’s helpful to review some tricks for dealing with inequalities. If an event is defined in terms of an inequality, the complement will be defined in terms of the opposite inequality: Both the direction and the inclusivity will be reversed, as shown in the table below.

Example 7.22

Using the formula for complements to compute probabilities.

• If you roll a standard 6-sided die, what is the probability that the result will be a number greater than one?
• If you roll two standard 6-sided dice, what is the probability that the sum will be 10 or less?
• If you flip a fair coin 3 times, what is the probability that at least one flip will come up tails?
• Here, the sample space is {1, 2, 3, 4, 5, 6}. It’s easy enough to see that the probability in question is 5 6 5 6 , because there are 5 outcomes that fall into the event “roll a number greater than 1.” Let’s also apply our new formula to find that probability. Since E E is defined using the inequality roll > 1 roll > 1 , then E ′ E ′ is defined using roll ≤ 1 roll ≤ 1 . Since there’s only one outcome (1) in E ′ E ′ , we have P ( E ′ ) = 1 6 P ( E ′ ) = 1 6 . Thus, P ( E ) = 1 − P ( E ′ ) = 5 6 P ( E ) = 1 − P ( E ′ ) = 5 6 .

Here, the event E E is defined by the inequality sum ≤ 10 sum ≤ 10 . Thus, E ′ E ′ is defined by sum > 10 sum > 10 . There are three outcomes in E ′ E ′ : two 11s and one 12. Thus, P ( E ) = 1 − P ( E ′ ) = 1 − 3 36 = 11 12 P ( E ) = 1 − P ( E ′ ) = 1 − 3 36 = 11 12 .

• In Example 7.15, we found the sample space for this experiment consisted of these equally likely outcomes: {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}. Our event E E is defined by T ≥ 1 T ≥ 1 , so E ′ E ′ is defined by T < 1 T < 1 . The only outcome in E ′ E ′ is the first one on the list, where zero tails are flipped. So, P ( E ) = 1 − P ( E ′ ) = 1 − 1 8 = 7 8 P ( E ) = 1 − P ( E ′ ) = 1 − 1 8 = 7 8 .

Your Turn 7.22

Check your understanding, section 7.5 exercises.

For the following exercises, use the following table of the top 15 players by number of plate appearances (PA) in the 2019 Major League Baseball season to assign empirical probabilities to the given events. A plate appearance is a batter’s opportunity to try to get a hit. The other columns are runs scored (R), hits (H), doubles (2B), triples (3B), home runs (HR), walks (BB), and strike outs (SO).

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3.1: Sample Spaces, Events, and Their Probabilities

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Learning Objectives

• To learn the concept of the sample space associated with a random experiment.
• To learn the concept of an event associated with a random experiment.
• To learn the concept of the probability of an event.

Sample Spaces and Events

Rolling an ordinary six-sided die is a familiar example of a random experiment , an action for which all possible outcomes can be listed, but for which the actual outcome on any given trial of the experiment cannot be predicted with certainty. In such a situation we wish to assign to each outcome, such as rolling a two, a number, called the probability of the outcome, that indicates how likely it is that the outcome will occur. Similarly, we would like to assign a probability to any event , or collection of outcomes, such as rolling an even number, which indicates how likely it is that the event will occur if the experiment is performed. This section provides a framework for discussing probability problems, using the terms just mentioned.

Definition: random experiment

A random experiment is a mechanism that produces a definite outcome that cannot be predicted with certainty. The sample space associated with a random experiment is the set of all possible outcomes. An event is a subset of the sample space.

Definition: Element and Occurrence

An event \(E\) is said to occur on a particular trial of the experiment if the outcome observed is an element of the set \(E\).

Example \(\PageIndex{1}\): Sample Space for a single coin

Construct a sample space for the experiment that consists of tossing a single coin.

The outcomes could be labeled \(h\) for heads and \(t\) for tails. Then the sample space is the set: \(S = \{h,t\}\)

Example \(\PageIndex{2}\): Sample Space for a single die

Construct a sample space for the experiment that consists of rolling a single die. Find the events that correspond to the phrases “an even number is rolled” and “a number greater than two is rolled.”

The outcomes could be labeled according to the number of dots on the top face of the die. Then the sample space is the set \(S = \{1,2,3,4,5,6\}\)

The outcomes that are even are \(2, 4,\; \; \text{and}\; \; 6\), so the event that corresponds to the phrase “an even number is rolled” is the set \(\{2,4,6\}\), which it is natural to denote by the letter \(E\). We write \(E=\{2,4,6\}\).

Similarly the event that corresponds to the phrase “a number greater than two is rolled” is the set \(T=\{3,4,5,6\}\), which we have denoted \(T\).

A graphical representation of a sample space and events is a Venn diagram , as shown in Figure \(\PageIndex{1}\). In general the sample space \(S\) is represented by a rectangle, outcomes by points within the rectangle, and events by ovals that enclose the outcomes that compose them.

Example \(\PageIndex{3}\): Sample Spaces for two coines

A random experiment consists of tossing two coins.

• Construct a sample space for the situation that the coins are indistinguishable, such as two brand new pennies.
• Construct a sample space for the situation that the coins are distinguishable, such as one a penny and the other a nickel.
• After the coins are tossed one sees either two heads, which could be labeled \(2h\), two tails, which could be labeled \(2t\), or coins that differ, which could be labeled \(d\) Thus a sample space is \(S=\{2h, 2t, d\}\).
• Since we can tell the coins apart, there are now two ways for the coins to differ: the penny heads and the nickel tails, or the penny tails and the nickel heads. We can label each outcome as a pair of letters, the first of which indicates how the penny landed and the second of which indicates how the nickel landed. A sample space is then \(S' = \{hh, ht, th, tt\}\).

A device that can be helpful in identifying all possible outcomes of a random experiment, particularly one that can be viewed as proceeding in stages, is what is called a tree diagram . It is described in the following example.

Example \(\PageIndex{4}\): Tree diagram

Construct a sample space that describes all three-child families according to the genders of the children with respect to birth order.

Two of the outcomes are “two boys then a girl,” which we might denote \(bbg\), and “a girl then two boys,” which we would denote \(gbb\).

Clearly there are many outcomes, and when we try to list all of them it could be difficult to be sure that we have found them all unless we proceed systematically. The tree diagram shown in Figure \(\PageIndex{2}\), gives a systematic approach.

The diagram was constructed as follows. There are two possibilities for the first child, boy or girl, so we draw two line segments coming out of a starting point, one ending in a \(b\) for “boy” and the other ending in a \(g\) for “girl.” For each of these two possibilities for the first child there are two possibilities for the second child, “boy” or “girl,” so from each of the \(b\) and \(g\) we draw two line segments, one segment ending in a \(b\) and one in a \(g\). For each of the four ending points now in the diagram there are two possibilities for the third child, so we repeat the process once more.

The line segments are called branches of the tree. The right ending point of each branch is called a node . The nodes on the extreme right are the final nodes ; to each one there corresponds an outcome, as shown in the figure.

From the tree it is easy to read off the eight outcomes of the experiment, so the sample space is, reading from the top to the bottom of the final nodes in the tree,

\[S=\{bbb,\; bbg,\; bgb,\; bgg,\; gbb,\; gbg,\; ggb,\; ggg\} \nonumber \]

Probability

Definition: probability.

The probability of an outcome \(e\) in a sample space \(S\) is a number \(P\) between \(1\) and \(0\) that measures the likelihood that \(e\) will occur on a single trial of the corresponding random experiment. The value \(P=0\) corresponds to the outcome \(e\) being impossible and the value \(P=1\) corresponds to the outcome \(e\) being certain.

Definition: probability of an event

The probability of an event \(A\) is the sum of the probabilities of the individual outcomes of which it is composed. It is denoted \(P(A)\).

The following formula expresses the content of the definition of the probability of an event:

If an event \(E\) is \(E=\{e_1,e_2,...,e_k\}\), then

\[P(E)=P(e_1)+P(e_2)+...+P(e_k) \nonumber \]

The following figure expresses the content of the definition of the probability of an event:

Since the whole sample space \(S\) is an event that is certain to occur, the sum of the probabilities of all the outcomes must be the number \(1\).

In ordinary language probabilities are frequently expressed as percentages. For example, we would say that there is a \(70\%\) chance of rain tomorrow, meaning that the probability of rain is \(0.70\). We will use this practice here, but in all the computational formulas that follow we will use the form \(0.70\) and not \(70\%\).

Example \(\PageIndex{5}\)

A coin is called “balanced” or “fair” if each side is equally likely to land up. Assign a probability to each outcome in the sample space for the experiment that consists of tossing a single fair coin.

With the outcomes labeled \(h\) for heads and \(t\) for tails, the sample space is the set

\[S=\{h,t\} \nonumber \]

Since the outcomes have the same probabilities, which must add up to \(1\), each outcome is assigned probability \(1/2\).

Example \(\PageIndex{6}\)

A die is called “balanced” or “fair” if each side is equally likely to land on top. Assign a probability to each outcome in the sample space for the experiment that consists of tossing a single fair die. Find the probabilities of the events \(E\): “an even number is rolled” and \(T\): “a number greater than two is rolled.”

With outcomes labeled according to the number of dots on the top face of the die, the sample space is the set

\[S=\{1,2,3,4,5,6\} \nonumber \]

Since there are six equally likely outcomes, which must add up to \(1\), each is assigned probability \(1/6\).

Since \(E = \{2,4,6\}\),

\[P(E) = \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{3}{6} = \dfrac{1}{2} \nonumber \]

Since \(T = \{3,4,5,6\}\),

\[P(T) = \dfrac{4}{6} = \dfrac{2}{3} \nonumber \]

Example \(\PageIndex{7}\)

Two fair coins are tossed. Find the probability that the coins match, i.e., either both land heads or both land tails.

In Example \(\PageIndex{3}\) we constructed the sample space \(S=\{2h,2t,d\}\) for the situation in which the coins are identical and the sample space \(S′=\{hh,ht,th,tt\}\) for the situation in which the two coins can be told apart.

The theory of probability does not tell us how to assign probabilities to the outcomes, only what to do with them once they are assigned. Specifically, using sample space \(S\), matching coins is the event \(M=\{2h, 2t\}\) which has probability \(P(2h)+P(2t)\). Using sample space \(S'\), matching coins is the event \(M'=\{hh, tt\}\), which has probability \(P(hh)+P(tt)\). In the physical world it should make no difference whether the coins are identical or not, and so we would like to assign probabilities to the outcomes so that the numbers \(P(M)\) and \(P(M')\) are the same and best match what we observe when actual physical experiments are performed with coins that seem to be fair. Actual experience suggests that the outcomes in S' are equally likely, so we assign to each probability \(\frac{1}{4}\), and then...

\[P(M') = P(hh) + P(tt) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \nonumber \]

Similarly, from experience appropriate choices for the outcomes in \(S\) are:

\[P(2h) = \frac{1}{4} \nonumber \]

\[P(2t) = \frac{1}{4} \nonumber \]

\[P(d) = \frac{1}{2} \nonumber \]

The previous three examples illustrate how probabilities can be computed simply by counting when the sample space consists of a finite number of equally likely outcomes. In some situations the individual outcomes of any sample space that represents the experiment are unavoidably unequally likely, in which case probabilities cannot be computed merely by counting, but the computational formula given in the definition of the probability of an event must be used.

Example \(\PageIndex{8}\)

The breakdown of the student body in a local high school according to race and ethnicity is \(51\%\) white, \(27\%\) black, \(11\%\) Hispanic, \(6\%\) Asian, and \(5\%\) for all others. A student is randomly selected from this high school. (To select “randomly” means that every student has the same chance of being selected.) Find the probabilities of the following events:

• \(B\): the student is black,
• \(M\): the student is minority (that is, not white),
• \(N\): the student is not black.

The experiment is the action of randomly selecting a student from the student population of the high school. An obvious sample space is \(S=\{w,b,h,a,o\}\). Since \(51\%\) of the students are white and all students have the same chance of being selected, \(P(w)=0.51\), and similarly for the other outcomes. This information is summarized in the following table:

\[\begin{array}{l|cccc}Outcome & w & b & h & a & o \\ Probability & 0.51 & 0.27 & 0.11 & 0.06 & 0.05\end{array} \nonumber \]

• Since \(B=\{b\},\; \; P(B)=P(b)=0.27\)
• Since \(M=\{b,h,a,o\},\; \; P(M)=P(b)+P(h)+P(a)+P(o)=0.27+0.11+0.06+0.05=0.49\)
• Since \(N=\{w,h,a,o\},\; \; P(N)=P(w)+P(h)+P(a)+P(o)=0.51+0.11+0.06+0.05=0.73\)

Example \(\PageIndex{9}\)

The student body in the high school considered in the last example may be broken down into ten categories as follows: \(25\%\) white male, \(26\%\) white female, \(12\%\) black male, \(15\%\) black female, 6% Hispanic male, \(5\%\) Hispanic female, \(3\%\) Asian male, \(3\%\) Asian female, \(1\%\) male of other minorities combined, and \(4\%\) female of other minorities combined. A student is randomly selected from this high school. Find the probabilities of the following events:

• \(B\): the student is black
• \(MF\): the student is a non-white female
• \(FN\): the student is female and is not black

Now the sample space is \(S=\{wm, bm, hm, am, om, wf, bf, hf, af, of\}\). The information given in the example can be summarized in the following table, called a two-way contingency table:

• Since \(B=\{bm, bf\},\; \; P(B)=P(bm)+P(bf)=0.12+0.15=0.27\)
• Since \(MF=\{bf, hf, af, of\},\; \; P(M)=P(bf)+P(hf)+P(af)+P(of)=0.15+0.05+0.03+0.04=0.27\)
• Since \(FN=\{wf, hf, af, of\},\; \; P(FN)=P(wf)+P(hf)+P(af)+P(of)=0.26+0.05+0.03+0.04=0.38​​​​​​\)

Key Takeaway

• The sample space of a random experiment is the collection of all possible outcomes.
• An event associated with a random experiment is a subset of the sample space.
• The probability of any outcome is a number between \(0\) and \(1\). The probabilities of all the outcomes add up to \(1\).
• The probability of any event \(A\) is the sum of the probabilities of the outcomes in \(A\).

Probability

How likely something is to happen.

Many events can't be predicted with total certainty. The best we can say is how likely they are to happen, using the idea of probability.

Tossing a Coin

When a coin is tossed, there are two possible outcomes:

Heads (H) or Tails (T)

• the probability of the coin landing H is ½
• the probability of the coin landing T is ½

Throwing Dice

When a single die is thrown, there are six possible outcomes: 1, 2, 3, 4, 5, 6 .

The probability of any one of them is 1 6

In general:

Probability of an event happening = Number of ways it can happen Total number of outcomes

Example: the chances of rolling a "4" with a die

Number of ways it can happen: 1 (there is only 1 face with a "4" on it)

Total number of outcomes: 6 (there are 6 faces altogether)

So the probability = 1 6

Example: there are 5 marbles in a bag: 4 are blue, and 1 is red. What is the probability that a blue marble gets picked?

Number of ways it can happen: 4 (there are 4 blues)

Total number of outcomes: 5 (there are 5 marbles in total)

So the probability = 4 5 = 0.8

Probability Line

We can show probability on a Probability Line :

Probability is always between 0 and 1

Probability is Just a Guide

Probability does not tell us exactly what will happen, it is just a guide

Example: toss a coin 100 times, how many Heads will come up?

Probability says that heads have a ½ chance, so we can expect 50 Heads .

But when we actually try it we might get 48 heads, or 55 heads ... or anything really, but in most cases it will be a number near 50.

Learn more at Probability Index .

Some words have special meaning in Probability:

Experiment : a repeatable procedure with a set of possible results.

Example: Throwing dice

We can throw the dice again and again, so it is repeatable.

The set of possible results from any single throw is {1, 2, 3, 4, 5, 6}

Outcome: A possible result.

Example: "6" is one of the outcomes of a throw of a die.

Trial: A single performance of an experiment.

Example: I conducted a coin toss experiment. After 4 trials I got these results:

Three trials had the outcome "Head", and one trial had the outcome "Tail"

Sample Space: all the possible outcomes of an experiment.

Example: choosing a card from a deck

There are 52 cards in a deck (not including Jokers)

So the Sample Space is all 52 possible cards : {Ace of Hearts, 2 of Hearts, etc... }

The Sample Space is made up of Sample Points:

Sample Point: just one of the possible outcomes

Example: Deck of Cards

• the 5 of Clubs is a sample point
• the King of Hearts is a sample point

"King" is not a sample point. There are 4 Kings, so that is 4 different sample points.

There are 6 different sample points in that sample space.

Event: one or more outcomes of an experiment

Example Events:

An event can be just one outcome:

• Getting a Tail when tossing a coin
• Rolling a "5"

An event can include more than one outcome:

• Choosing a "King" from a deck of cards (any of the 4 Kings)
• Rolling an "even number" (2, 4 or 6)

Hey, let's use those words, so you get used to them:

Example: Alex wants to see how many times a "double" comes up when throwing 2 dice.

The Sample Space is all possible Outcomes (36 Sample Points):

{1,1} {1,2} {1,3} {1,4} ... ... ... {6,3} {6,4} {6,5} {6,6}

The Event Alex is looking for is a "double", where both dice have the same number. It is made up of these 6 Sample Points :

{1,1} {2,2} {3,3} {4,4} {5,5} and {6,6}

These are Alex's Results:

 After 100 Trials , Alex has 19 "double" Events ... is that close to what you would expect?

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Keyboard Shortcuts

Section 1: introduction to probability.

In the lessons that follow, and throughout the rest of this course, we'll be learning all about the basics of probability — its properties, how it behaves, and how to calculate a probability. To do so, we'll be simultaneously working through this section, that is, Section 1, and the first chapter in the Hogg and Tanis textbook (10th edition).

7.6 Probability with Permutations and Combinations

Learning objectives.

After completing this section, you should be able to:

• Calculate probabilities with permutations.
• Calculate probabilities with combinations.

In our earlier discussion of theoretical probabilities, the first step we took was to write out the sample space for the experiment in question. For many experiments, that method just isn’t practical. For example, we might want to find the probability of drawing a particular 5-card poker hand. Since there are 52 cards in a deck and the order of cards doesn’t matter, the sample space for this experiment has 52 C 5 = 2,598,960 52 C 5 = 2,598,960 possible 5-card hands. Even if we had the patience and space to write them all out, sorting through the results to find the outcomes that fall in our event would be just as tedious.

Luckily, the formula for theoretical probabilities doesn’t require us to know every outcome in the sample space; we just need to know how many outcomes there are. In this section, we’ll apply the techniques we learned earlier in the chapter ( The Multiplication Rule for Counting , permutations, and combinations) to compute probabilities.

Using Permutations to Compute Probabilities

Recall that we can use permutations to count how many ways there are to put a number of items from a list in order. If we’re looking at an experiment whose sample space looks like an ordered list, then permutations can help us to find the right probabilities.

Example 7.23

• In horse racing, an exacta bet is one where the player tries to predict the top two finishers in particular race in order. If there are 9 horses in a race, and a player decided to make an exacta bet at random, what is the probability that they win?
• You are in a club with 10 people, 3 of whom are close friends of yours. If the officers of this club are chosen at random, what is the probability that you are named president and one of your friends is named vice president?
• A bag contains slips of paper with letters written on them as follows: A, A, B, B, B, C, C, D, D, D, D, E. If you draw 3 slips, what is the probability that the letters will spell out (in order) the word BAD?
• Since order matters for this situation, we’ll use permutations. How many different exacta bets can be made? Since there are 9 horses and we must select 2 in order, we know there are 9 P 2 = 56 9 P 2 = 56 possible outcomes. That’s the size of our sample space, so it will go in the denominator of the probability. Since only one of those outcomes is a winner, the numerator of the probability is 1. So, the probability of randomly selecting the winning exacta bet is 1 56 1 56 .
• There are 10 people in the club, and 2 will be chosen to be officers. Since the order matters, there are 10 P 2 = 90 10 P 2 = 90 different ways to select officers. Next, we must figure out how many outcomes are in our event. We’ll use the Multiplication Rule for Counting to find that number. There is only 1 choice for president in our event, and there are 3 choices for vice president. So, there are 1 × 3 = 3 1 × 3 = 3 outcomes in the event. Thus, the probability that you will serve as president with one of your friends as vice president is 3 90 = 1 30 3 90 = 1 30 .
• There are 12 slips of paper in the bag, and 3 will be drawn. So, there are 12 P 3 = 1320 12 P 3 = 1320 possible outcomes. Now, we’ll compute the number of outcomes in our event. The first letter drawn must be a B, and there are 3 of those. Next must come an A (2 of those) and then a D (4 of those). Thus, there are 3 × 2 × 4 = 24 3 × 2 × 4 = 24 outcomes in our event. So, the probability that the letters drawn spell out the word BAD is 24 1320 = 1 55 24 1320 = 1 55 .

Your Turn 7.23

Combinations to computer probabilities.

If the sample space of our experiment is one in which order doesn’t matter, then we can use combinations to find the number of outcomes in that sample space.

Example 7.24

Using combinations to compute probabilities.

• Palmetto Cash 5 is a game offered by the South Carolina Education Lottery. Players choose 5 numbers from the whole numbers between 1 and 38 (inclusive); the player wins the jackpot of $100,000 if the randomizer selects those numbers in any order. If you buy one ticket for this game, what is the probability that you win the top prize by choosing all 5 winning numbers?
• There’s a second prize in the Palmetto Cash 5 game that a player wins if 4 of the player's 5 numbers are among the 5 winning numbers. What’s the probability of winning the second prize?
• Scrabble is a word-building board game. Players make hands of 7 letters by selecting tiles with single letters printed on them blindly from a bag (2 tiles have nothing printed on them; these blanks can stand for any letter). Players use the letters in their hands to spell out words on the board. Initially, there are 100 tiles in the bag. Of those, 44 are (or could be) vowels (9 As, 12 Es, 9 Is, 8 Os, 4 Us, and 2 blanks; we’ll treat Y as a consonant). What is the probability that your initial hand has no vowels?
• There are 38 numbers to choose from, and the order of the 5 we pick doesn’t matter. So, there are 38 C 5 = 501 , 492 38 C 5 = 501 , 492 outcomes in the sample space. Only one outcome is in our winning event, so the probability of winning is 1 501 , 492 1 501 , 492 .
• As in part 1 of this example,, there are 501,492 outcomes in the sample space. The tricky part here is figuring out how many outcomes are in our event. To qualify, the outcome must contain 4 of the 5 winning numbers, plus one losing number. There are 5 C 4 = 5 5 C 4 = 5 ways to choose the 4 winning numbers, and there are 38 − 5 = 33 38 − 5 = 33 losing numbers. So, using the Multiplication Rule for Counting, there are 5 × 33 = 165 5 × 33 = 165 outcomes in our event. Thus, the probability of winning the second prize is 165 501 , 492 = 55 167 , 164 165 501 , 492 = 55 167 , 164 , which is about 0.00033.
• The number of possible starting hands is 100 C 7 = 16 , 007 , 560 , 800 100 C 7 = 16 , 007 , 560 , 800 . There are 100 − 44 = 56 100 − 44 = 56 consonants in the bag, so the number of all-consonant hands is 56 C 7 = 231 , 917 , 400 56 C 7 = 231 , 917 , 400 . Thus, the probability of drawing all consonants is 231 , 917 , 40 16 , 007 , 560 , 800 = 32 , 139 2 , 425 , 388 ≈ 0.0145 231 , 917 , 40 16 , 007 , 560 , 800 = 32 , 139 2 , 425 , 388 ≈ 0.0145 .

Your Turn 7.24

Check your understanding, section 7.6 exercises.

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Probability Calculator

Table of contents

With the probability calculator, you can investigate the relationships of likelihood between two separate events . For example, if the chance of A happening is 50%, and the same for B, what are the chances of both happening, only one happening, at least one happening, or neither happening, and so on.

Our probability calculator gives you six scenarios, plus 6 more when you enter in how many times the "die is cast", so to speak. As long as you know how to find the probability of individual events, it will save you a lot of time.

Reading on below, you'll:

• Discover how to use the probability calculator properly;
• Check how to find the probability of single events;
• Read about multiple examples of probability usage, including conditional probability formulas;
• Study the difference between a theoretical and empirical probability; and
• Increase your knowledge about the relationship between probability and statistics.

Did you come here specifically to check your odds of winning a bet or hitting the jackpot? Our odds calculator and lottery calculator will assist you!

How to find the probability of events? – probability definition

The basic definition of probability is the ratio of all favorable results to the number of all possible outcomes.

Allowed values of a single probability vary from 0 to 1 , so it's also convenient to write probabilities as percentages. The probability of a single event can be expressed as such:

• The probability of A : P(A) ,
• The probability of B : P(B) ,
• The probability of + : P(+) ,
• The probability of ♥ : P(♥) , etc.

Let's take a look at an example with multi-colored balls. We have a bag filled with orange, green, and yellow balls. Our event A is picking a random ball out of the bag . We can define Ω as a complete set of balls. The probability of event Ω , which means picking any ball, is naturally 1. In fact, a sum of all possible events in a given set is always equal to 1 .

Now let's look at something more challenging – what's the likelihood of picking an orange ball? To answer this question, you have to find the number of all orange marbles and divide it by the number of all balls in the bag. You can do it for any color, e.g., yellow, and you'll undoubtedly notice that the more balls in a particular color, the higher the probability of picking it out of the bag if the process is totally random.

Check out our probability calculator 3 events and conditional probability calculator for determining the chances of multiple events.

We can define a complementary event , written as Ā or A' , which means not A . In our example, the probability of picking out NOT an orange ball is evaluated as a number of all non-orange ones divided by all marbles. The sum P(A) + P(Ā) is always 1 because there is no other option like half of a ball or a semi-orange one.

Now, try to find the probability of getting a blue ball. No matter how hard you try, you will fail because there is not even one in the bag, so the result is equal to 0.

We use intuitive calculations of probability all the time. Knowing how to quantify likelihood is essential for statistical analysis. It allows you to measure this otherwise nebulous concept called "probability". Furthermore, given a discrete dataset, the relative frequency for each value is synonymous with the probability of their occurrence.

Are you looking for something slightly different? Take a look at our post-test probability calculator . 🎲

How to use the probability calculator?

To make the most of our calculator, you'll need to take the following steps:

1. Define the problem you want to solve.

Your problem needs to be condensed into two independent events.

2. Find the probability of each event.

Now, when you know how to estimate the likelihood of a single event, you only need to perform the task and obtain all of the necessary values.

3. Type the percentage probability of each event in the corresponding fields.

Once they're in, the probability calculator will immediately populate with the exact likelihood of 6 different scenarios:

• Both events will happen;
• At least one of the events will happen;
• Exactly one of the events will happen;
• Neither of the events will happen;
• Only the first event won't happen; and
• Only the second event won't happen

You can also choose to see all of the above. Additionally, the calculator can also show the probability of six more scenarios, given a certain number of trials:

• A always occurring;
• A never occurring;
• A occurring at least once;
• B always occurring;
• B never occurring; and
• B occurring at least once.

You can change the number of trials and any other field in the calculator, and the other fields will automatically adjust themselves. This feature saves a ton of time if you want to find out, for example, what the probability of event B would need to become in order to make the likelihood of both occurring 50%.

If the set of possible choices is extremely large and only a few outcomes are successful, the resulting probability is tiny, like P(A) = 0.0001 . It's convenient to use scientific notation in order not to mix up the number of zeros.

Conditional probability

One of the most crucial considerations in the world of probabilities is whether the events are dependent or not. Two events are independent if the occurrence of the first one doesn't affect the likelihood of the occurrence of the second one . For example, if we roll a perfectly balanced standard cubic die, the possibility of getting a two ⚁ is equal to 1/6 (the same as getting a four ⚃ or any other number).

Let's say you have two dice rolls, and you get a five ⚄ in the first one. If you ask yourself what's the probability of getting a two ⚁ in the second turn, the answer is 1/6 once again because of the independence of events.

The way of thinking, as well as calculations, change if one of the events interrupts the whole system. This time we're talking about conditional probability .

Let's say we have 10 different numbered billiard balls, from ➀ to ➉. You choose a random ball, so the probability of getting the ➆ is precisely 1/10 . Suppose you picked the three ➂ and removed it from the game . Then you ask yourself, once again, what is the chance of getting the seven ➆. The situation changed because there is one ball with ➆ out of nine possibilities, which means that the probability is 1/9 now. In other words, the question can be asked: "What's the probability of picking ➆, IF the first ball was ➂?"

Let's look at another example: imagine that you are going to sit an exam in statistics. You know from your older colleagues that it's challenging, and the probability that you pass in the first term is 0.5 ( 18 out of 36 students passed last year). Then let's ask yourself a question: "What's the probability of passing IF you've already studied the topic?" 20 people admitted to reviewing their notes at least once before the exam, and 16 out of those succeeded, which means that the answer to the last question is 0.8 . This result indicates that this additional condition really matters if we want to find whether studying changes anything or not.

If you still don't feel the concept of conditional probability, let's try with another example: you have to drive from city X to city Y by car. The distance between them is about 150 miles. On the full tank, you can usually go up to 400 miles. If you don't know the fuel level, you can estimate the likelihood of successfully reaching the destination without refueling. And what if somebody has already filled the tank? Now you're almost sure that you can make it unless other issues prevent it.

Conditional probability formula

The formal expression of conditional probability, which can be denoted as P(A|B) , P(A/B) or P B (A) , can be calculated as:

P(A|B) = P(A∩B) / P(B) ,

where P(B) is the probability of an event B , and P(A∩B) is the joint of both events. On the other hand, we can estimate the intersection of two events if we know one of the conditional probabilities:

• P(A∩B) = P(A|B) * P(B) or
• P(A∩B) = P(B|A) * P(A) .

It's better to understand the concept of conditional probability formula with tree diagrams. We ask students in a class if they like Math and Physics. An event M denotes the percentage that enjoys Math, and P the same for Physics:

There is a famous theorem that connects conditional probabilities of two events. It's named Bayes' theorem , and the formula is as follows:

P(A|B) = P(B|A) * P(A) / P(B)

You can ask a question: "What is the probability of A given B if I know the likelihood of B given A ?". This theorem sometimes provides surprising and unintuitive results. The most commonly described examples are drug testing and illness detection, which has a lot in common with the relative risk of disease in the population. Let's stick to the second one. In a group of 1000 people, 10 of them have a rare disease. Everybody had a test, which shows the actual result in 95% of cases. So now we want to find the probability of a person being ill if their test result is positive.

Without thinking, you may predict, by intuition, that the result should be around 90% , right? Let's make some calculations and estimate the correct answer.

• We will use a notation: H – healthy, I – ill, + – test positive, - – test negative.
• Rewrite information from the text above in a way of probabilities: P(H) = 0.99 , P(I) = 0.01 , P(+|I) = 0.95 , P(-|I) = 0.05 , P(+|H) = 0.05 , P(-|H) = 0.95 .
• Work out the total probability of a test to be positive: P(+) = P(+|I) * P(I) + P(+|H) * P(H) = 0.95 * 0.01 + 0.05 * 0.99 = 0.059 .
• Use the Bayes' theorem to find the conditional probability P(I|+) = P(+|I) * P(I) / P(+) = 0.95 * 0.01 / 0.059 = 0.161 .

Hmm... it isn't that high, is it? It turns out that this kind of paradox appears if there is a significant imbalance between the number of healthy and ill people , or in general, between two distinct groups. If the result is positive, it's always worth repeating the test to make an appropriate diagnosis.

Probability distribution and cumulative distribution function

We can distinguish between two kinds of probability distributions, depending on whether the random variables are discrete or continuous.

A discrete probability distribution describes the likelihood of the occurrence of countable, distinct events. One of the examples is binomial probability, which takes into account the probability of some kind of success in multiple turns, e.g., while tossing a coin. In contrast, in the Pascal distribution (also known as negative binomial) the fixed number of successes is given, and you want to estimate the total number of trials.

The Poisson distribution is another discrete probability distribution and is actually a particular case of binomial one, which you can calculate with our Poisson distribution calculator . The probability mass function can be interpreted as another definition of discrete probability distribution – it assigns a given value to any separate number. The geometric distribution is an excellent example of using the probability mass function.

A continuous probability distribution holds information about uncountable events. It's impossible to predict the likelihood of a single event (like in a discrete one), but rather that we can find the event in some range of variables. The normal distribution is one of the best-known continuous distribution. It describes a bunch of properties within any population, e.g., the height of adult people or the IQ dissemination. The function that describes the probability of seeing a result from a given range of values is called the probability density function .

If you are more advanced in probability theory and calculations, you definitely have to deal with SMp(x) distribution , which takes into account the combination of several discrete and continuous probability functions.

For each probability distribution, we can construct the cumulative distribution function (CDF) . It tells you what the probability is that some variable will take the value less than or equal to a given number .

Let's say you participate in a general knowledge quiz. The competition consists of 100 questions, and you earn 1 point for a correct answer, whereas for the wrong one, there are no points. Many people have already finished, and out of the results, we can obtain a probability distribution. Rules state that only 20% best participants receive awards, so you wonder how well you should score to be one of the winners. If you look at the graph, you can divide it so that 80% of the area below is on the left side and 20% of the results are on the right of the desired score. What you are actually looking for is a left-tailed p-value.

However, there is also another way to find it if we use a cumulative distribution function – just find the value 80% on the axis of abscissa and the corresponding number of points without calculating anything!

Theoretical vs experimental probability

Almost every example described above takes into account the theoretical probability. So a question arises: what's the difference between theoretical and experimental (also known as empirical) probability? The formal definition of theoretical probability is the ratio between the number of favorable outcomes to the number of every possible outcome . It relies on the given information, logical reasoning and tells us what we should expect from an experiment .

Just look at bags with colorful balls once again. There are 42 marbles in total, and 18 of them are orange. The game consists of picking a random ball from the bag and putting it back, so there are always 42 balls inside. Applying the probability definition, we can quickly estimate it as 18/42 , or simplifying the fraction, 3/7 . It means that if we pick 14 balls, there should be 6 orange ones.

On the other hand, the experimental probability tells us precisely what happened when we perform an experiment instead of what should happen. It is based on the ratio of the number of successful and the number of all trials . Let's stick with the same example – pick a random marble from the bag and repeat the procedure 13 more times. Suppose you get 8 orange balls in 14 trials. This result means that the empirical probability is 8/14 or 4/7 .

As you can see, your outcome differs from the theoretical one. It's nothing strange because when you try to reiterate this game over and over, sometimes, you will pick more, and sometimes you will get less, and sometimes you will pick exactly the number predicted theoretically. If you sum up all results, you should notice that the overall probability gets closer and closer to the theoretical probability . If not, then we can suspect that picking a ball from the bag isn't entirely random, e.g., the balls of different colors have unequal sizes, so you can distinguish them without having to look.

Probability and statistics

Both statistics and probability are the branches of mathematics and deal with the relationship of the occurrence of events . However, everyone should be aware of the differences which make them two distinct areas.

Probability is generally a theoretical field of math, and it investigates the consequences of mathematical definitions and theorems . In contrast, statistics is usually a practical application of mathematics in everyday situations and tries to attribute sense and understanding of the observations in the real world .

Probability predicts the possibility of events to happen , whereas statistics is basically analyzing the frequency of the occurrence of past ones and creates a model based on the acquired knowledge .

Imagine a probabilist playing a card game, which relies on choosing a random card from the whole deck, knowing that only spades win with predefined odds ratio. Assuming that the deck is complete and the choice is entirely random and equitable, they deduce that the probability is equal to ¼ and can make a bet.

A statistician is going to observe the game for a while first to check if, in fact, the game is fair. After verifying (with acceptable approximation) that the game is worth playing, then he will ask the probabilist what he should do to win the most.

Statistics within a large group of people – probability sampling

You've undoubtedly seen some election preference polls, and you may have wondered how they may be quite so precise in comparison to final scores, even if the number of people asked is way lower than the total population – this is the time when probability sampling takes place .

The underlying assumption, which is the basic idea of sampling, is that the volunteers are chosen randomly with a previously defined probability. We can distinguish between multiple kinds of sampling methods:

• Simple random sampling
• Cluster random sampling
• Systematic sampling
• Probability-proportional-to-size sampling
• Stratified random sampling
• Minimax sampling
• Accidental sampling
• Quota sampling
• Voluntary sampling
• Panel sampling
• Snowball sampling
• Line-intercept sampling
• Theoretical sampling

Each of these methods has its advantages and drawbacks, but most of them are satisfactory. Significant benefits of probability sampling are time-saving, and cost-effectiveness since a limited number of people needs to be surveyed. The simplicity of this procedure doesn't require any expertise and can be performed without any thorough preparation.

Practical application of probability theory

As you could have already realized, there are a lot of areas where the theory of probability is applicable. Most of them are games with a high random factor, like rolling dice or picking one colored ball out of 10 different colors, or many card games. Lotteries and gambling are the kinds of games that extensively use the concept of probability and the general lack of knowledge about it. Of course, somebody wins from time to time, but the likelihood that the person will be you is extremely small .

Probability theory is also used in many different types of problems. Especially when talking about investments, it is also worth considering the risk to choose the most appropriate option.

Our White Christmas calculator uses historical data and probability knowledge to predict the occurrence of snow cover for many cities during Christmas.

How do I calculate the probability of A and B?

If A and B are independent events , then you can multiply their probabilities together to get the probability of both A and B happening. For example, if the probability of A is 20% (0.2) and the probability of B is 30% (0.3) , the probability of both happening is 0.2 × 0.3 = 0.06 = 6% .

How do I calculate conditional probability?

To compute the conditional probability of A under B :

• Determine the probability of B , i.e., P(B) .
• Determine the probability of A and B , i.e., P(A∩B) .
• Divide the result from Step 2 by that of Step 1.
• That's it! The formula reads: P(A|B) = P(A∩B) / P(B) .

What's the probability of rolling 2 sixes?

If you are using fair dice, the probability of rolling two sixes will be 1/6 × 1/6 = 1/36 = 0.027 = 2.7% . That means it takes 36 dice rolls to expect rolling 2 sixes at least once, though there's no guarantee when it comes to probability.

How do I convert odds to percentage?

Convert the odds to a decimal number, then multiply by 100. For example, if the odds are 1 in 9, that's 1/9 = 0.1111 in decimal form. Then multiply by 100 to get 11.11% .

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ol{padding-top:0px;}.css-4okk7a ul:not(:first-child),.css-4okk7a ol:not(:first-child){padding-top:4px;} Probabilities of single events

Probability of A: P(A)

Probability of B: P(B)

Which probability do you want to see?

Type of probability

Probability of the intersection of A and B.

Probabilities for a series of events

The probability of

...when trying

If you want to find the conditional probability, check out our Bayes' Theorem Calculator !

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AP®︎/College Statistics

Course: ap®︎/college statistics   >   unit 7.

• Conditional probability and independence
• Conditional probability with Bayes' Theorem

Conditional probability using two-way tables

• Calculate conditional probability
• Conditional probability tree diagram example
• Tree diagrams and conditional probability

• Your answer should be
• an integer, like 6 ‍  
• a simplified proper fraction, like 3 / 5 ‍  
• a simplified improper fraction, like 7 / 4 ‍  
• a mixed number, like 1   3 / 4 ‍  
• an exact decimal, like 0.75 ‍  
• a multiple of pi, like 12   pi ‍   or 2 / 3   pi ‍  
• (Choice A)   True A True
• (Choice B)   False B False
• (Choice A)   About 62 % ‍   of females chose invisibility as their superpower. A About 62 % ‍   of females chose invisibility as their superpower.
• (Choice B)   About 62 % ‍   of people who chose invisibility as their superpower were female. B About 62 % ‍   of people who chose invisibility as their superpower were female.

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• Math Article

Probability And Statistics

Probability And Statistics are the two important concepts in Maths. Probability is all about chance. Whereas statistics is more about how we handle various data using different techniques.  It helps to represent complicated data in a very easy and understandable way. Statistics and probability are usually introduced in Class 10, Class 11 and Class 12 students are preparing for school exams and competitive examinations. The introduction of these fundamentals is briefly given in your academic books and notes. The statistic has a huge application nowadays in data science professions. The professionals use the stats and do the predictions of the business. It helps them to predict the future profit or loss attained by the company.

Table of contents:

• Probability
• Probability Topics
• Statistics Topics

Solved Examples

What is probability.

Probability denotes the possibility of the outcome of any random event. The meaning of this term is to check the extent to which any event is likely to happen. For example, when we flip a coin in the air, what is the possibility of getting a head ? The answer to this question is based on the number of possible outcomes. Here the possibility is either head or tail will be the outcome. So, the probability of a head to come as a result is 1/2.

The probability is the measure of the likelihood of an event to happen. It measures the certainty of the event. The formula for probability is given by;

P(E) = Number of Favourable Outcomes/Number of total outcomes

P(E) = n(E)/n(S)

n(E) = Number of event favourable to event E

n(S) = Total number of outcomes

What is Statistics?

Statistics is the study of the collection, analysis, interpretation, presentation, and organization of data. It is a method of collecting and summarising the data. This has many applications from a small scale to large scale. Whether it is the study of the population of the country or its economy, stats are used for all such data analysis.

Statistics has a huge scope in many fields such as sociology, psychology, geology, weather forecasting, etc. The data collected here for analysis could be quantitative or qualitative. Quantitative data are also of two types such as: discrete and continuous. Discrete data has a fixed value whereas continuous data is not a fixed data but has a range. There are many terms and formulas used in this concept. See the below table to understand them .

Terms Used in Probability and Statistics

There are various terms utilised in the probability and statistics concepts, Such as:

Random Experiment

• Sample Sample
• Random variables

Expected Value

• Independence

Let us discuss these terms one by one.

An experiment whose result cannot be predicted, until it is noticed is called a random experiment. For example, when we throw a dice randomly, the result is uncertain to us. We can get any output between 1 to 6. Hence, this experiment is random.

Sample Space

A sample space is the set of all possible results or outcomes of a random experiment. Suppose, if we have thrown a dice, randomly, then the sample space for this experiment will be all possible outcomes of throwing a dice, such as;

Sample Space = { 1,2,3,4,5,6}

Random Variables

The variables which denote the possible outcomes of a random experiment are called random variables. They are of two types:

• Discrete Random Variables
• Continuous Random Variables

Discrete random variables take only those distinct values which are countable. Whereas continuous random variables could take an infinite number of possible values.

Independent Event

When the probability of occurrence of one event has no impact on the probability of another event, then both the events are termed as independent of each other. For example, if you flip a coin and at the same time you throw a dice, the probability of getting a ‘head’ is independent of the probability of getting a 6 in dice.

Mean of a random variable is the average of the random values of the possible outcomes of a random experiment. In simple terms, it is the expectation of the possible outcomes of the random experiment, repeated again and again or n number of times. It is also called the expectation of a random variable.

Expected value is the mean of a random variable. It is the assumed value which is considered for a random experiment.  It is also called expectation, mathematical expectation or first moment. For example, if we roll a dice having six faces, then the expected value will be the average value of all the possible outcomes, i.e. 3.5.

Basically, the variance tells us how the values of the random variable are spread around the mean value. It specifies the distribution of the sample space across the mean.

List of Probability Topics

Basic probability topics are :

List of Statistical Topics

Basic Statistics topics are:

Probability and Statistics Formulas

Probability Formulas : For two events A and B:

Statistics Formulas : Some important formulas are listed below:

Let x be an item given and n is the total number of items.

Here are some examples based on the concepts of statistics and probability to understand better. Students can practice more questions based on these solved examples to excel in the topic. Also, make use of the formulas given in this article in the above section to solve problems based on them.

Example 1 : Find the mean and mode of the following data: 2, 3, 5, 6, 10, 6, 12, 6, 3, 4.

Total Count: 10

Sum of all the numbers: 2+3+5+6+10+6+12+6+3+7=60

Mean = (sum of all the numbers)/(Total number of items)

Mean = 60/10 = 6

Again, Number 6 is occurring for 3 times, therefore Mode = 6. Answer

Example 2: A bucket contains 5 blue, 4 green and 5 red balls. Sudheer is asked to pick 2 balls randomly from the bucket without replacement and then one more ball is to be picked. What is the probability he picked 2 green balls and 1 blue ball?

Solution : Total number of balls = 14

Probability of drawing

1 green ball = 4/14

another green ball = 3/13

1 blue ball = 5/12

Probability of picking 2 green balls and 1 blue ball = 4/14 * 3/13 * 5/12 = 5/182.

Example 3 : What is the probability that Ram will choose a marble at random and that it is not black if the bowl contains 3 red, 2 black and 5 green marbles.

Solution : Total number of marble = 10

Red and Green marbles = 8

Find the number of marbles that are not black and divide by the total number of marbles.

So P(not black) = (number of red or green marbles)/(total number of marbles)

Example 4: Find the mean of the following data:

55, 36, 95, 73, 60, 42, 25, 78, 75, 62

Solution: Given,

55 36 95 73 60 42 25 78 75 62

Sum of observations = 55 + 36 + 95 + 73 + 60 + 42 + 25 + 78 + 75 + 62 = 601

Number of observations = 10

Mean = 601/10 = 60.1

Example 5: Find the median and mode of the following marks (out of 10) obtained by 20 students:

4, 6, 5, 9, 3, 2, 7, 7, 6, 5, 4, 9, 10, 10, 3, 4, 7, 6, 9, 9

Ascending order: 2, 3, 3, 4, 4, 4, 5, 5, 6, 6, 6, 7, 7, 7, 9, 9, 9, 9, 10, 10

Number of observations = n = 20

Median = (10th  + 11th observation)/2

= (6 + 6)/2

Most frequent observations = 9

Hence, the mode is 9.

Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin!

Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz

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