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Conditional Probability Questions

Conditional probability questions with solutions are given here for students to practice and understand the concept of conditional probability. By conditional probability, we mean the possibility of happening an event after the occurrence of an event already. For example, a dice is rolled, and the outcome is an odd number, now we have to find the probability of the outcome being a prime number, then,

A ≡ Odd outcome for rolling a dice

B ≡ Outcome being a prime number

Clearly, A already happened thereafter; we have to find the probability of happening B.

Thus, P(B|A) = P(A ∩ B)/P(A), where P(A) ≠ 0 and

P(B|A) = probability of occurrence of B given A already happened

P(A ∩ B) = probability of occurrence of A and B together

P(A) = probability of occurrence of A.

Learn more about conditional probability .

Conditional Probability Questions with Solutions

Let us solve some questions based on conditional probability with detailed solutions.

Question 1:

Ten numbered cards are there from 1 to 15, and two cards a chosen at random such that the sum of the numbers on both the cards is even. Find the probability that the chosen cards are odd-numbered.

Let, A ≡ event of selecting two odd-numbered cards

B ≡ event of selecting cards whose sum is even.

n(B) = number of ways of choosing two numbers whose sum is even = 8 C 2 + 7 C 2 .

n(A ∩ B) = number of ways of choosing odd-numbered cards such that their sum is even.

Now, P(A|B) = P(A ∩ B)/P(B) = n(A ∩ B)/n(B)

= 8 C 2 / ( 8 C 2 + 7 C 2 ) = 4/7.

Question 2:

Let E and F are events of a experiment such that P(E) = 3/10 P(F) = ½ and P(F|E) = ⅖. Find the value of (i) P(E ∩ F) (ii) P(E|F) (iii) P(E U F)

We know that P(A|B) = P(A ∩ B)/P(B) ⇒ P(A ∩ B) = P(A|B).P(B)

∴ P(E ∩ F) = P(F|E).P(E) =⅖ × 3/10 = 3/25

(ii) P(E|F) = P(E ∩ F)/P(F) = (3/25) ÷ (½) = 6/25

(iii) P(E U F) = P(E) + P(F) – P(E ∩ F) = 3/10 + ½ – 3/25 = 17/25.

Question 3:

The probability of a student passing in science is ⅘ and the of the student passing in both science and maths is ½. What is the probability of that student passing in maths knowing that he passed in science?

Let A ≡ event of passing in science

B ≡ event of passing in maths

Given, P(B) = ⅘ and P(A ∩ B) = ½

Then, probability of passing maths after passing in science = P(B|A) = P(A ∩ B)/P(A)

= ½ ÷ ⅘ = ⅝

∴ the probability of passing in maths is ⅝.

Question 4:

In a survey among few people, 60% read Hindi newspaper, 40% read English newspaper and 20% read both. If a person is chosen at random and if he already reads English newspaper find the probability that he also reads Hindi newspaper.

Let there be 100 people in the survey, then

Number of people read Hindi newspaper = n(A) = 60

Number of people read English newspaper = n(B) = 40

Number of people read both = n(A ∩ B) = 20

Probability of the person reading Hindi newspaper when he already reads English newspaper is given by –

P(A|B) = n(A ∩ B)/n(B) = 20/40 = ½.

• Bayes Theorem
• Probability for Class 12
• Basic Probability Formulas
• Mutually Exclusive Events

Question 5:

A fair coin is tossed twice such that E: event of having both head and tail, and F: event of having atmost one tail. Find P(E), P(F) and P(E|F)

The sample space S = { HH, HT, TH, TT}

E = {HT, TH}

F = {HH, HT, TH}

E ∩ F = {HT, TH}

P(E) = 2/4 = ½

P(E ∩ F) = 2/4 = ½

P(E|F) = P(E ∩ F)/P(F) = ½ ÷ ¾ = ⅔.

Question 6:

In a class, 40% of the students like Mathematics and 25% of students like Physics and 15% like both the subjects. One student select at random, find the probability that he likes Physics if it is known that he likes Mathematics.

Let there be 100 students, then,

Number of students like Mathematics = n(A) = 40

Number of students like Physics = n(B) = 25

Number of students like both Mathematics and Physics = n(A ∩ B) = 15

Now, the probability that the student likes Physics if it is known that he likes Mathematics is given by –

P(B|A) = n(A ∩ B)/n(A) = 15/40 = ⅜.

Question 7:

Two dice are rolled, if it is known that atleast one of the dice always shows 4, find the probability that the numbers appeared on the dice have a sum 8.

A: one of the outcomes is always 4

B: sum of the outcomes is 8

Then, A = {(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6)}

B{(4, 4), (5, 3), (3, 5), (6, 2), (2, 6)}

n(A) = 11, n(B) = 5, n(A ∩ B) = 1

P(B|A) = n(A ∩ B)/n(A) = 1/11.

Question 8:

A bag contains 3 red and 7 black balls. Two balls are drawn at randon without replacement. If the second ball is red, what is the probability that the first ball is also red?

Let A: event of selecting a red ball in first draw

B: event of selecting a red ball in second draw

P(A ∩ B) = P(selecting both red balls) = 3/10 × 2/9 = 1/15

P(B) = P(selecting a red ball in second draw) = P(red ball and rad ball or black ball and red ball)

= P(red ball and red ball) + P(black ball and red ball)

= 3/10 × 2/9 + 7/10 × 3/9 = 3/10

∴ P(A|B) = P(A ∩ B)/P(B) = 1/15 ÷ 3/10 = 2/9.

Question 9:

If a family has two children, what is the conditional probability that both are girls if there is atleast one girl?

Let A: both being girls

B: Atleast one girl

n(A ∩ B) = 1

P(A|B) = n(A ∩ B)/n(B) = ⅓.

Question 10:

A dice and a coin are tossed simultaneously. Find the probability of obtaining a 6, given that a head came up.

Let A: six coming with a heads

B: coin shows a head

A ={(6, H))

B = {(1, H), (2, H), (3, H), (4, H), (5, H), (6, H)}

n(A ∩ B) = 1 and n(B) = 6

∴ Probability of getting a six when there is a head is given by –

P(A|B) = n(A ∩ B)/n(B) = ⅙.

Practice Questions on Conditional Probability

1. If P(A) = 2P(B) = 4/13 and P(A|B) = ⅔, find the value of P(A U B).

2. In a survey among few people, 60% read Hindi newspaper, 40% read English newspaper and 20% read both. If a person is chosen at random and if he already reads Hindi newspaper find the probability that he also reads English newspaper.

3. A fair coin is tossed twice such that E: event of having both head and tail, and F: event of having atmost one head. Find P(E ∩ F) and P(F|E).

4. In a class, 60% of the students like Mathematics and 35% of students like Physics and 25% like both the subjects. One student select at random, find the probability that he likes Physics if it is known that he likes Mathematics.

5. Two dice are rolled, if it is known that the second dice always shows 4, find the probability that the numbers appeared on the dice have a sum 6.

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Conditional Probability

Practice Conditional Probabilities

Conditional Probability

How to handle Dependent Events

Life is full of random events! You need to get a "feel" for them to be a smart and successful person.

Independent Events

Events can be " Independent ", meaning each event is not affected by any other events.

Example: Tossing a coin.

Each toss of a coin is a perfect isolated thing.

What it did in the past will not affect the current toss.

The chance is simply 1-in-2, or 50%, just like ANY toss of the coin.

So each toss is an Independent Event .

Dependent Events

But events can also be "dependent" ... which means they can be affected by previous events ...

Example: Marbles in a Bag

2 blue and 3 red marbles are in a bag.

What are the chances of getting a blue marble?

The chance is 2 in 5

But after taking one out the chances change!

So the next time:

This is because we are removing marbles from the bag.

So the next event depends on what happened in the previous event, and is called dependent .

Replacement

Note: if we replace the marbles in the bag each time, then the chances do not change and the events are independent :

• With Replacement: the events are Independent (the chances don't change)
• Without Replacement: the events are Dependent (the chances change)

Dependent events are what we look at here.

Tree Diagram

A Tree Diagram is a wonderful way to picture what is going on, so let's build one for our marbles example.

There is a 2/5 chance of pulling out a Blue marble, and a 3/5 chance for Red:

We can go one step further and see what happens when we pick a second marble:

If a blue marble was selected first there is now a 1/4 chance of getting a blue marble and a 3/4 chance of getting a red marble.

If a red marble was selected first there is now a 2/4 chance of getting a blue marble and a 2/4 chance of getting a red marble.

Now we can answer questions like "What are the chances of drawing 2 blue marbles?"

Answer: it is a 2/5 chance followed by a 1/4 chance :

Did you see how we multiplied the chances? And got 1/10 as a result.

The chances of drawing 2 blue marbles is 1/10

We love notation in mathematics! It means we can then use the power of algebra to play around with the ideas. So here is the notation for probability:

P(A) means "Probability Of Event A"

In our marbles example Event A is "get a Blue Marble first" with a probability of 2/5:

And Event B is "get a Blue Marble second" ... but for that we have 2 choices:

• If we got a Blue Marble first the chance is now 1/4
• If we got a Red Marble first the chance is now 2/4

So we have to say which one we want , and use the symbol "|" to mean "given":

P(B|A) means "Event B given Event A"

In other words, event A has already happened, now what is the chance of event B?

P(B|A) is also called the "Conditional Probability" of B given A.

And in our case:

P(B|A) = 1/4

So the probability of getting 2 blue marbles is:

And we write it as

"Probability of event A and event B equals the probability of event A times the probability of event B given event A "

Let's do the next example using only notation:

Example: Drawing 2 Kings from a Deck

Event A is drawing a King first, and Event B is drawing a King second.

For the first card the chance of drawing a King is 4 out of 52 (there are 4 Kings in a deck of 52 cards):

P(A) = 4/52

But after removing a King from the deck the probability of the 2nd card drawn is less likely to be a King (only 3 of the 51 cards left are Kings):

P(B|A) = 3/51

P(A and B) = P(A) x P(B|A) =(4/52)x (3/51) = 12/2652 = 1/221

So the chance of getting 2 Kings is 1 in 221, or about 0.5%

Finding Hidden Data

Using Algebra we can also "change the subject" of the formula, like this:

And we have another useful formula:

"The probability of event B given event A equals the probability of event A and event B divided by the probability of event A "

Example: Ice Cream

70% of your friends like Chocolate, and 35% like Chocolate AND like Strawberry.

What percent of those who like Chocolate also like Strawberry?

P(Strawberry|Chocolate) = P(Chocolate and Strawberry) / P(Chocolate)

50% of your friends who like Chocolate also like Strawberry

Big Example: Soccer Game

You are off to soccer, and want to be the Goalkeeper, but that depends who is the Coach today:

• with Coach Sam the probability of being Goalkeeper is 0.5
• with Coach Alex the probability of being Goalkeeper is 0.3

Sam is Coach more often ... about 6 out of every 10 games (a probability of 0.6 ).

So, what is the probability you will be a Goalkeeper today?

Let's build a tree diagram . First we show the two possible coaches: Sam or Alex:

The probability of getting Sam is 0.6, so the probability of Alex must be 0.4 (together the probability is 1)

Now, if you get Sam, there is 0.5 probability of being Goalie (and 0.5 of not being Goalie):

If you get Alex, there is 0.3 probability of being Goalie (and 0.7 not):

The tree diagram is complete, now let's calculate the overall probabilities. Remember that:

P(A and B) = P(A) x P(B|A)

Here is how to do it for the "Sam, Yes" branch:

(When we take the 0.6 chance of Sam being coach times the 0.5 chance that Sam will let you be Goalkeeper we end up with an 0.3 chance.)

But we are not done yet! We haven't included Alex as Coach:

With 0.4 chance of Alex as Coach, followed by the 0.3 chance gives 0.12

And the two "Yes" branches of the tree together make:

0.3 + 0.12 = 0.42 probability of being a Goalkeeper today

(That is a 42% chance)

One final step: complete the calculations and make sure they add to 1:

0.3 + 0.3 + 0.12 + 0.28 = 1

Yes, they add to 1 , so that looks right.

Friends and Random Numbers

Here is another quite different example of Conditional Probability.

4 friends (Alex, Blake, Chris and Dusty) each choose a random number between 1 and 5. What is the chance that any of them chose the same number?

Let's add our friends one at a time ...

First, what is the chance that Alex and Blake have the same number?

Blake compares his number to Alex's number. There is a 1 in 5 chance of a match.

As a tree diagram :

Note: "Yes" and "No" together  makes 1 (1/5 + 4/5 = 5/5 = 1)

Now, let's include Chris ...

But there are now two cases to consider:

• If Alex and Blake did match, then Chris has only one number to compare to.
• But if Alex and Blake did not match then Chris has two numbers to compare to.

And we get this:

For the top line (Alex and Blake did match) we already have a match (a chance of 1/5).

But for the "Alex and Blake did not match" there is now a 2/5 chance of Chris matching (because Chris gets to match his number against both Alex and Blake).

And we can work out the combined chance by multiplying the chances it took to get there:

Following the "No, Yes" path ... there is a 4/5 chance of No, followed by a 2/5 chance of Yes:

Following the "No, No" path ... there is a 4/5 chance of No, followed by a 3/5 chance of No:

Also notice that when we add all chances together we still get 1 (a good check that we haven't made a mistake):

(5/25) + (8/25) + (12/25) = 25/25 = 1

Now what happens when we include Dusty?

It is the same idea, just more of it:

OK, that is all 4 friends, and the "Yes" chances together make 101/125:

Answer: 101/125

But here is something interesting ... if we follow the "No" path we can skip all the other calculations and make our life easier:

The chances of not matching are:

(4/5) × (3/5) × (2/5) = 24/125

So the chances of matching are:

1 - (24/125) = 101/125

(And we didn't really need a tree diagram for that!)

And that is a popular trick in probability:

It is often easier to work out the "No" case (and subtract from 1 for the "Yes" case)

(This idea is shown in more detail at Shared Birthdays .)

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11.4.1: Conditional Probability (Exercises)

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SECTION 8.4 PROBLEM SET: CONDITIONAL PROBABILITY

Questions 1 - 4: Do these problems using the conditional probability formula: \(P(A | B)=\frac{P(A \cap B)}{P(B)}\).

Questions 5 - 8 refer to the following: The table shows the distribution of Democratic and Republican U.S. Senators by gender in the 114 th Congress as of January 2015.

Use this table to determine the following probabilities:

Do the following conditional probability problems.

At a college, 72% of courses have final exams and 46% of courses require research papers. 32% of courses have both a research paper and a final exam. Let \(F\) be the event that a course has a final exam and \(R\) be the event that a course requires a research paper.

Consider a family of three children. Find the following probabilities.

Questions 21 - 26 refer to the following: The table shows highest attained educational status for a sample of US residents age 25 or over:

Conditional Probability Worksheets

What Is Conditional Probability? The probability that is defined as the possibility for the likelihood of the following event based on the result of the previous event. Conditional probability is found out by multiplying the probability of the previous into the probability of the succeeding event. For example: You have event A that represents that there is a 20% chance of rain today. You have event B that represents that you have to go outside and has the probability of 60%. The conditional probability will give you the outcome based on the relationship between these two events. That is the probability that represents both, you need to go outside, and it is raining. Independent events are events that are not affected by one another. For example, tossing a coin. In a coin toss, each toss is an independent event and an isolated outcome. Events can also be dependent. That is, both the events are affected by each other. The next event is affected by the probability of the previous event. For example, 2 blue and 3 red balls are in the bag. The probability of getting a blue ball is 2 in 5. But if we got a red ball before, then the probability of getting a blue is 2 in 4.

Basic Lesson

Introduces the application of conditional probability with playing cards. One card is drawn from a standard deck of cards. Find the probability of drawing a club or spade card?

Intermediate Lesson

This lesson focuses on conditional probability within random numbers. A number is selected from 1 to 15 at random. What is the probability of getting a number that is a multiple of 5?

Independent Practice 1

Students practice with 20 conditional probability problems. The answers can be found below. In a middle school of 200 students, 100 are seniors, 50 students take part in a play, and 30 students are seniors and are also taking part in play. What is the probability that a randomly chosen student is a senior and is taking part in play?

Independent Practice 2

Another 20 conditional probability problems. The answers can be found below. If the student selected is an 8th grade student, what is the probability that the student prefers Science or History?

Homework Worksheet

Reviews all skills in the unit. A great take home sheet. Also provides a practice problem.

10 problems that test Conditional Probability skills.

Homework and Quiz Answer Key

Answers for the homework and quiz.

Answers for the lesson and practice sheets.

What Are The Odds?

A statistician was stopped because she had a bomb in her carry-on. The official said, "Why would you want to blow up a plane?" The woman replied, "I wasn't going to use it, but the odds of there being 2 bombs on one plane is 1 in 100,000."

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7.2: Conditional Probability

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Often it is required to compute the probability of an event given that another event has occurred.

Example \(\PageIndex{1}\)

What is the probability that two cards drawn at random from a deck of playing cards will both be aces?

It might seem that you could use the formula for the probability of two independent events and simply multiply \(\dfrac{4}{52} \cdot \dfrac{4}{52}=\dfrac{1}{169}\). This would be incorrect, however, because the two events are not independent. Once the first card is drawn, there are only 51 cards remaining in the deck. If the first card drawn is an ace, then the probability that the second card is also an ace would be \(\dfrac{3}{51}\) because there would only be three aces left in the deck.

Once the first card chosen is an ace, the probability that the second card chosen is also an ace is called the conditional probability of drawing an ace. In this case, the "condition" is that the first card is an ace. Symbolically, we write this as:

\(P(\) ace on second draw \(\mid\) an ace on the first draw).

The vertical bar "|" is read as "given," so the above expression is short for "The probability that an ace is drawn on the second draw given that an ace was drawn on the first draw." What is this probability? After an ace is drawn on the first draw, there are 3 aces out of 51 total cards left. This means that the conditional probability of drawing an ace after one ace has already been drawn is \(\dfrac{3}{51}=\dfrac{1}{17}\).

Thus, the probability of both cards being aces is \(\dfrac{4}{52} \cdot \dfrac{3}{51}=\dfrac{12}{2652}=\dfrac{1}{221}\).

Definition: Conditional Probability

The probability that event B occurs, given that event A has happened, is represented as P(B | A) This is read as “the probability of B given A”

Example \(\PageIndex{2}\)

Find the probability that a die rolled shows a 6, given that a flipped coin shows a head.

These are two independent events, so the probability of the die rolling a 6 is \(\dfrac{1}{6}\), regardless of the result of the coin flip.

Example \(\PageIndex{3}\)

The table below shows the number of survey subjects who have received and not received a speeding ticket in the last year, and the color of their cars. Find the probability that a randomly chosen person:

a) Has a speeding ticket given they have a red car b) Has a red car given they have a speeding ticket

a) Since we know the person has a red car, we are only considering the 150 people in the first row of the table. Of those, 15 have a speeding ticket, so

\[ \mathrm{P}(\text { ticket } \mid \text { red car })=\dfrac{15}{150}=\dfrac{1}{10}=0.1 \]

b) Since we know the person has a speeding ticket, we are only considering the 60 people in the first column of the table. Of those, 15 have a red car, so

\[ \mathrm{P}(\text { red car } \mid \text { ticket })=\dfrac{15}{60}=\dfrac{1}{4}=0.25 \text {. } \]

Notice from the last example that P(B | A) is not equal to P(A | B).

These kinds of conditional probabilities are what insurance companies use to determine your insurance rates. They look at the conditional probability of you having an accident, given your age, your car, your car color, your driving history, etc., and price your policy based on that likelihood.

P(A and B) for dependent events

If Events A and B are dependent (not independent), then P(A and B) = P(A) · P(B | A)

Example \(\PageIndex{4}\)

If you pull 2 cards out of a deck, what is the probability that both are spades?

The probability that the first card is a spade is \(\dfrac{13}{52}\).

The probability that the second card is a spade, given the first was a spade, is \(\dfrac{12}{51}\), since there is one less spade in the deck, and one less total cards.

The probability that both cards are spades is \(\dfrac{13}{52} \cdot \dfrac{12}{51}=\dfrac{156}{2652} \approx 0.0588\)

You Try It \(\PageIndex{1}\)

If you draw 2 cards from a standard deck, what is the probability that both cards are red kings?

Example \(\PageIndex{5}\)

A home pregnancy test was given to women, then pregnancy was verified through blood tests. The following table shows the home pregnancy test results. Find

a) P(not pregnant | positive test result) b) P(positive test result | not pregnant)

a) Since we know the test result was positive, we're limited to the 75 women in the first column, of which 5 were not pregnant.

\[P(\) not pregnant \(\mid\) positive test result \()=\dfrac{5}{75} \approx 0.067\].

b) Since we know the woman is not pregnant, we are limited to the 19 women in the second row, of which 5 had a positive test.

\[ P \text { (positive test result } \mid \text { not pregnant })=\dfrac{5}{19} \approx 0.263 \]

The second result is what is usually called a false positive: A positive result when the woman is not actually pregnant.

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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

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Video transcript

6.4. Conditional Probability

Conditional probability.

Suppose you and a friend wish to play a game that involves choosing a single card from a well-shuffled deck. Your friend deals you one card, face down, from the deck and offers you the following deal: if the card is a king, he will pay you $5, otherwise, you pay him $1. Should you play the game?

You reason in the following manner. Since there are four kings in the deck, the probability of obtaining a king is 4/52 or 1/13. And, the probability of not obtaining a king is 12/13. This implies that the ratio of your winning to losing is 1 to 12, while the payoff ratio is only $1 to $5. Therefore, you determine that you should not play.

Now consider the following scenario. While your friend was dealing the card, you happened to get a glance of it and noticed that the card was a face card. Should you, now, play the game?

Since there are 12 face cards in the deck, the total elements in the sample space are no longer 52, but just 12. This means the chance of obtaining a king is 4/12 or 1/3. So your chance of winning is 1/3 and of losing 2/3. This makes your winning to losing ratio 1 to 2 which fares much better with the payoff ratio of $1 to $5. This time, you determine that you should play.

In the second part of the above example, we were finding the probability of obtaining a king knowing that a face card had shown. This is an example of conditional probability . Whenever we are finding the probability of an event E under the condition that another event F has happened, we are finding conditional probability.

The symbol P ( E |  F ) denotes the problem of finding the probability of E given that F has occurred. We read P ( E |  F ) as “the probability of E , given F .”

Example 6.4.1

Conditional probability formula

For two events E and F , the probability of E given F is:

Example 6.4.2

Example 6.4.3

The events M , F , P , and D are self explanatory. Find the following probabilities:

Example 6.4.4

Example 6.4.5

Example 6.4.6

Using the conditional probability formula, we get:

Substituting:

Example 6.4.7

Let event E be that the family has two boys and a girl, and let F be the probability that the family has at least two boys. We want to find P ( E | F ). We list the sample space along with the events E and F:

Example 6.4.8

Practice questions

1. A die is rolled. Use the conditional probability formula to find the conditional probability that it shows a three if it is known that an odd number has shown.

2. The following table shows the distribution of coffee drinkers by gender:

Use the table to determine the following probabilities:

3. In the Occupational and Public Health program at Ryerson University, 60% of the students pass Biostatistics, 70% pass Environmental Health Law, and 30% pass both of these courses. If a student is selected at random, find the following conditional probabilities:

a . They pass Biostatistics given that they passed Law

b. They pass Law given that they passed Biostatistics

4. Consider a family of three children. What is the probability of the family having children of both sexes given that the first born child is a boy?

6. A survey of drivers was conducted to determine the number of speeding tickets received among males and females. The data are displayed in the table below.

a . P (0 speeding tickets)

b. P (F | 1 speeding ticket)

c. P (M | at least 2 speeding tickets)

Mathematics for Public and Occupational Health Professionals Copyright © 2019 by Ian Young is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License , except where otherwise noted.