conditional probability homework 3

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 .)

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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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.

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4.3: Conditional Probability

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• Page ID 16347

• Kathryn Kozak
• Coconino Community College

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Suppose you want to figure out if you should buy a new car. When you first go and look, you find two cars that you like the most. In your mind they are equal, and so each has a 50% chance that you will pick it. Then you start to look at the reviews of the cars and realize that the first car has had 40% of them needing to be repaired in the first year, while the second car only has 10% of the cars needing to be repaired in the first year. You could use this information to help you decide which car you want to actually purchase. Both cars no longer have a 50% chance of being the car you choose. You could actually calculate the probability you will buy each car, which is a conditional probability. You probably wouldn’t do this, but it gives you an example of what a conditional probability is.

Conditional probabilities are probabilities calculated after information is given. This is where you want to find the probability of event A happening after you know that event B has happened. If you know that B has happened, then you don’t need to consider the rest of the sample space. You only need the outcomes that make up event B. Event B becomes the new sample space, which is called the restricted sample space , R. If you always write a restricted sample space when doing conditional probabilities and use this as your sample space, you will have no trouble with conditional probabilities. The notation for conditional probabilities is \(P(A, \text { given } B)=P(A | B)\). The event following the vertical line is always the restricted sample space.

Example \(\PageIndex{1}\) conditional probabilities

• Suppose you roll two dice. What is the probability of getting a sum of 5, given that the first die is a 2?
• Suppose you roll two dice. What is the probability of getting a sum of 7, given the first die is a 4?
• Suppose you roll two dice. What is the probability of getting the second die a 2, given the sum is a 9?
• Suppose you pick a card from a deck. What is the probability of getting a Spade, given that the card is a Jack?
• Suppose you pick a card from a deck. What is the probability of getting an Ace, given the card is a Queen?

a. Since you know that the first die is a 2, then this is your restricted sample space, so

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

Out of this restricted sample space, the way to get a sum of 5 is {(2,3)}. Thus

\(P(\text { sum of } 5 | \text { the first die is a } 2)=\dfrac{1}{6}\)

b. Since you know that the first die is a 4, this is your restricted sample space, so

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

Out of this restricted sample space, the way to get a sum of 7 is {(4,3)}. Thus

\(P(\text { sum of } 7 | \text { the first die is a } 4)=\dfrac{1}{6}\)

c. Since you know the sum is a 9, this is your restricted sample space, so

R = {(3,6), (4,5), (5,4), (6,3)}

Out of this restricted sample space there is no way to get the second die a 2. Thus

\(P(\text { second die is a } 2 | \text { sum is } 9)=0\)

d. Since you know that the card is a Jack, this is your restricted sample space, so

R = {JS, JC, JD, JH}

Out of this restricted sample space, the way to get a Spade is {JS}. Thus

\(P(\text { Spade } | \mathrm{Jack})=\dfrac{1}{4}\)

e. on: Since you know that the card is a Queen, then this is your restricted sample space, so

R = {QS, QC, QD, QH}

Out of this restricted sample space, there is no way to get an Ace, thus

\(P(\text { Ace | Queen })=0\)

If you look at the results of Example \(\PageIndex{7}\) part d and Example \(\PageIndex{1}\) part b, you will notice that you get the same answer. This means that knowing that the first die is a 4 did not change the probability that the sum is a 7. This added knowledge did not help you in any way. It is as if that information was not given at all. However, if you compare Example \(\PageIndex{7}\) part b and Example \(\PageIndex{1}\) part a, you will notice that they are not the same answer. In this case, knowing that the first die is a 2 did change the probability of getting a sum of 5. In the first case, the events sum of 7 and first die is a 4 are called independent events . In the second case, the events sum of 5 and first die is a 2 are called dependent events .

Events A and B are considered independent events if the fact that one event happens does not change the probability of the other event happening. In other words, events A and B are independent if the fact that B has happened does not affect the probability of event A happening and the fact that A has happened does not affect the probability of event B happening. Otherwise, the two events are dependent. In symbols, A and B are independent if

\(P(A | B)=P(A) \text { or } P(B | A)=P(B)\)

Example \(\PageIndex{2}\) independent events

• Suppose you roll two dice. Are the events “sum of 7” and “first die is a 3” independent?
• Suppose you roll two dice. Are the events “sum of 6” and “first die is a 4” independent?
• Suppose you pick a card from a deck. Are the events “Jack” and “Spade” independent?
• Suppose you pick a card from a deck. Are the events “Heart” and “Red” card independent?
• Suppose you have two children via separate births. Are the events “the first is a boy” and “the second is a girl” independent?
• Suppose you flip a coin 50 times and get a head every time, what is the probability of getting a head on the next flip?

a. To determine if they are independent, you need to see if \(P(A | B)=P(A)\). It doesn’t matter which event is A or B, so just assign one as A and one as B.

Let A = sum of 7 = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)} and B = first die is a 3 = {(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)} \(P(A | B)\) means that you assume that B has happened. The restricted sample space is B,

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

In this restricted sample space, the way for A to happen is {(3,4)}, so

\(P(A | B)=\dfrac{1}{6}\)

The \(P(A)=\dfrac{6}{36}=\dfrac{1}{6}\)

\(P(A | B)=P(A)\) Thus “sum of 7” and “first die is a 3” are independent events.

b. To determine if they are independent, you need to see if \(P(A | B)=P(A)\). It doesn’t matter which event is A or B, so just assign one as A and one as B.

Let A = sum of 6 = {(1,5), (2,4), (3,3), (4,2), (5,1)} and B = first die is a 4 = {(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)}, so

\(P(A)=\dfrac{5}{36}\)

For \(P(A | B)\), the restricted sample space is B,

In this restricted sample space, the way for A to happen is {(4,2)}, so

\(P(A | B)=\dfrac{1}{6}\).

In this case, “sum of 6” and “first die is a 4” are dependent since \(P(A | B) \neq P(A)\).

c. To determine if they are independent, you need to see if \(P(A | B)=P(A)\). It doesn’t matter which event is A or B, so just assign one as A and one as B.

Let A = Jack = {JS, JC, JD, JH} and B = Spade {2S, 3S, 4S, 5S, 6S, 7S, 8S, 9S, 10S, JS, QS, KS, AS}

\(P(A)=\dfrac{4}{52}=\dfrac{1}{13}\)

R = {2S, 3S, 4S, 5S, 6S, 7S, 8S, 9S, 10S, JS, QS, KS, AS}

In this restricted sample space, the way A happens is {JS}, so

\(P(A | B)=\dfrac{1}{13}\)

In this case, “Jack” and “Spade” are independent since \(P(A | B)=P(A)\).

d. To determine if they are independent, you need to see if \(P(A | B)=P(A)\). It doesn’t matter which event is A or B, so just assign one as A and one as B.

Let A = Heart = {2H, 3H, 4H, 5H, 6H, 7H, 8H, 9H, 10H, JH, QH, KH, AH} and B = Red card = {2D, 3D, 4D, 5D, 6D, 7D, 8D, 9D, 10D, JD, QD, KD, AD, 2H, 3H, 4H, 5H, 6H, 7H, 8H, 9H, 10H, JH, QH, KH, AH}, so

\(P(A)=\dfrac{13}{52}=\dfrac{1}{4}\)

R = {2D, 3D, 4D, 5D, 6D, 7D, 8D, 9D, 10D, JD, QD, KD, AD, 2H, 3H, 4H, 5H, 6H, 7H, 8H, 9H, 10H, JH, QH, KH, AH}

In this restricted sample space, the way A can happen is 13,

\(P(A | B)=\dfrac{13}{26}=\dfrac{1}{2}\).

In this case, “Heart” and “Red” card are dependent, since \(P(A | B) \neq P(A)\).

e. In this case, you actually don’t need to do any calculations. The gender of one child does not affect the gender of the second child, the events are independent.

f. Since one flip of the coin does not affect the next flip (the coin does not remember what it did the time before), the probability of getting a head on the next flip is still one-half.

Multiplication Rule:

Two more useful formulas: If two events are dependent, then \(P(A \text { and } B)=P(A) * P(B | A)\)

If two events are independent, then \(P(A \text { and } B)=P(A)^{*} P(B)\)

If you solve the first equation for \(P(B | A)\), you obtain \(P(B | A)=\dfrac{P(A \text { and } B)}{P(A)}\), which is a formula to calculate a conditional probability. However, it is easier to find a conditional probability by using the restricted sample space and counting unless the sample space is large.

Example \(\PageIndex{3}\) Multiplication rule

• Suppose you pick three cards from a deck, what is the probability that they are all Queens if the cards are not replaced after they are picked?
• Suppose you pick three cards from a deck, what is the probability that they are all Queens if the cards are replaced after they are picked and before the next card is picked?

a. This sample space is too large to write out, so using the multiplication rule makes sense. Since the cards are not replaced, then the probability will change for the second and third cards. They are dependent events. This means that on the second draw there is one less Queen and one less card, and on the third draw there are two less Queens and 2 less cards.

P (3 Queens)= P (Q on 1st and Q on 2nd and Q on 3rd)

= P (Q on 1st)* P (Q on 2nd|Q on 1st)* P (Q on 3rd|1st and 2nd Q)

\(=\dfrac{4}{52} * \dfrac{3}{51} * \dfrac{2}{50}\)

\(=\dfrac{24}{132600}\)

b. Again, the sample space is too large to write out, so using the multiplication rule makes sense. Since the cards are put back, one draw has no affect on the next draw and they are all independent.

P (3 Queens)= P (Queen on 1st and Queen on 2nd and Queen on 3rd)

= P (Queen on 1st)* P (Queen on 2nd)* P (Queen on 3rd)

\(=\dfrac{4}{52} * \dfrac{4}{52} * \dfrac{4}{52}\)

\(=\left(\dfrac{4}{52}\right)^{3}\)

\(=\dfrac{64}{140608}\)

Example \(\PageIndex{4}\) application problem

The World Health Organization (WHO) keeps track of how many incidents of leprosy there are in the world. Using the WHO regions and the World Banks income groups, one can ask if an income level and a WHO region are dependent on each other in terms of predicting where the disease is. Data on leprosy cases in different countries was collected for the year 2011 and a summary is presented in Example \(\PageIndex{1}\) ("Leprosy: Number of," 2013).

• Find the probability that a person with leprosy is from the Americas.
• Find the probability that a person with leprosy is from a high-income country.
• Find the probability that a person with leprosy is from the Americas and a high-income country.
• Find the probability that a person with leprosy is from a high-income country, given they are from the Americas.
• Find the probability that a person with leprosy is from a low-income country.
• Find the probability that a person with leprosy is from Africa.
• Find the probability that a person with leprosy is from Africa and a low-income country.
• Find the probability that a person with leprosy is from Africa, given they are from a low-income country.
• Are the events that a person with leprosy is from “Africa” and “low-income country” independent events? Why or why not?
• Are the events that a person with leprosy is from “Americas” and “high-income country” independent events? Why or why not?

a. There are 36817 cases of leprosy in the Americas out of 222,545 cases worldwide. So,

\(P(\text { Americas })=\dfrac{36817}{222545} \approx 0.165\)

There is about a 16.5% chance that a person with leprosy lives in a country in the Americas.

b. There are 264 cases of leprosy in high-income countries out of 222,545 cases worldwide. So,

\(P(\text { high-income })=\dfrac{264}{222545} \approx 0.0001\)

There is about a 0.1% chance that a person with leprosy lives in a high-income country.

c. There are 174 cases of leprosy in countries in a high-income country in the Americas out the 222,545 cases worldwide. So,

\(P(\text { Americas and high-income })=\dfrac{174}{222545} 0.0008\)

There is about a 0.08% chance that a person with leprosy lives in a high-income country in the Americas.

d. In this case you know that the person is in the Americas. You don’t need to consider people from Easter Mediterranean, Europe, Western Pacific, Africa, and South-east Asia. You only need to look at the row with Americas at the start. In that row, look to see how many leprosy cases there are from a high-income country. There are 174 countries out of the 36,817 leprosy cases in the Americas. So,

\(P(\text { high-income } | \text { Americas })=\dfrac{174}{36817} \approx 0.0047\)

There is 0.47% chance that a person with leprosy is from a high-income country given that they are from the Americas.

e. There are 27,923 cases of leprosy in low-income countries out of the 222,545 leprosy cases worldwide. So,

\(P(\text { low-income })=\dfrac{27923}{222545} \approx 0.125\)

There is a 12.5% chance that a person with leprosy is from a low-income country.

f. There are 17,953 cases of leprosy in Africa out of 222,545 leprosy cases worldwide. So,

\(P(\text { Africa })=\dfrac{17953}{222545} \approx 0.081\)

There is an 8.1% chance that a person with leprosy is from Africa.

g. There are 15,928 cases of leprosy in low-income countries in Africa out of all the 222,545 leprosy cases worldwide. So,

\(P(\text { Africa and low-income })=\dfrac{15928}{222545} \approx 0.072\)

There is a 7.2% chance that a person with leprosy is from a low-income country in Africa.

h. In this case you know that the person with leprosy is from low-income country. You don’t need to include the high income, upper-middle income, and lowermiddle income country. You only need to consider the column headed by lowincome. In that column, there are 15,928 cases of leprosy in Africa out of the 27,923 cases of leprosy in low-income countries. So,

\(P(\text { Africa | low-income })=\dfrac{15928}{27923} \approx 0.570\)

There is a 57.0% chance that a person with leprosy is from Africa, given that they are from a low-income country.

i. In order for these events to be independent, either \(P(\text { Africa } | \text { low-income })=P(\text { Africa })\) or \(P(\text { low-income } | \text { Africa })=P(\text { low-income })\) have to be true. Part (h) showed \(P(\text { Africa | low-income }) \approx 0.570\) and part (f) showed \(P(\text { Africa }) \approx 0.081\). Since these are not equal, then these two events are dependent.

j. In order for these events to be independent, either \(P(\text { Americas } | \text { high-income })=P(\text { Americas })\) or \(P(\text { high-income |} \text { Americas })=P(\text { high-income })\) have to be true. Part (d) showed \(P(\text { high-income } | \text { Americas }) \approx 0.0047\) and part (b) showed \(P(\text { high-income }) \approx 0.001\). Since these are not equal, then these two events are dependent.

A big deal has been made about the difference between dependent and independent events while calculating the probability of and compound events. You must multiply the probability of the first event with the conditional probability of the second event.

Why do you care? You need to calculate probabilities when you are performing sampling, as you will learn later. But here is a simplification that can make the calculations a lot easier: when the sample size is very small compared to the population size, you can assume that the conditional probabilities just don't change very much over the sample.

For example, consider acceptance sampling. Suppose there is a big population of parts delivered to you factory, say 12,000 parts. Suppose there are 85 defective parts in the population. You decide to randomly select ten parts, and reject the shipment. What is the probability of rejecting the shipment?

There are many different ways you could reject the shipment. For example, maybe the first three parts are good, one is bad, and the rest are good. Or all ten parts could be bad, or maybe the first five. So many ways to reject! But there is only one way that you’d accept the shipment: if all ten parts are good. That would happen if the first part is good, and the second part is good, and the third part is good, and so on. Since the probability of the second part being good is (slightly) dependent on whether the first part was good, technically you should take this into consideration when you calculate the probability that all ten are good.

The probability of getting the first sampled part good is \(\dfrac{12000-85}{12000}=\dfrac{11915}{12000}\). So the probability that all ten being good is \(\dfrac{11915}{12000} * \dfrac{11914}{11999} * \dfrac{11913}{11998} * \ldots * \dfrac{11906}{11991} \approx 93.1357 \%\). If instead you assume that the probability doesn’t change much, you get \(\left(\dfrac{11915}{12000}\right)^{10} \approx 93.1382 \%\). So as you can see, there is not much difference. So here is the rule: if the sample is very small compared to the size of the population, then you can assume that the probabilities are independent, even though they aren’t technically. By the way, the probability of rejecting the shipment is \(1-0.9314=0.0686=6.86 \%\).

Exercise \(\PageIndex{1}\)

• Are owning a refrigerator and owning a car independent events? Why or why not?
• Are owning a computer or tablet and paying for Internet service independent events? Why or why not?
• Are passing your statistics class and passing your biology class independent events? Why or why not?
• Are owning a bike and owning a car independent events? Why or why not?
• What is the probability of picking a Jack given that the card is a face card?
• What is the probability of picking a heart given that the card is a three?
• What is the probability of picking a red card given that the card is an ace?
• Are the events Jack and face card independent events? Why or why not?
• Are the events red card and ace independent events? Why or why not?
• What is the probability that the sum is 6 given that the first die is a 5?
• What is the probability that the first die is a 3 given that the sum is 11?
• What is the probability that the sum is 7 given that the fist die is a 2?
• Are the two events sum of 6 and first die is a 5 independent events? Why or why not?
• Are the two events sum of 7 and first die is a 2 independent events? Why or why not?
• You flip a coin four times. What is the probability that all four of them are heads?
• You flip a coin six times. What is the probability that all six of them are heads?
• You pick three cards from a deck with replacing the card each time before picking the next card. What is the probability that all three cards are kings?
• You pick three cards from a deck without replacing a card before picking the next card. What is the probability that all three cards are kings?

Table \(\PageIndex{2}\) : Surviving the Titanic a. What is the probability that a survivor was female? b. What is the probability that a survivor was in the 1st class? c. What is the probability that a survivor was a female given that the person was in 1st class? d. What is the probability that a survivor was a female and in the 1st class? e. What is the probability that a survivor was a female or in the 1st class? f. Are the events survivor is a female and survivor is in 1st class mutually exclusive? Why or why not? g. Are the events survivor is a female and survivor is in 1st class independent? Why or why not?

Table \(\PageIndex{3}\) : Dolphin Activity a. What is the probability that a dolphin group is partaking in travel? b. What is the probability that a dolphin group is around in the morning? c. What is the probability that a dolphin group is partaking in travel given that it is morning? d. What is the probability that a dolphin group is around in the morning given that it is partaking in socializing? e. What is the probability that a dolphin group is around in the afternoon given that it is partaking in feeding? f. What is the probability that a dolphin group is around in the afternoon and is partaking in feeding? g. What is the probability that a dolphin group is around in the afternoon or is partaking in feeding? h. Are the events dolphin group around in the afternoon and dolphin group feeding mutually exclusive events? Why or why not? i. Are the events dolphin group around in the morning and dolphin group partaking in travel independent events? Why or why not?

1. Independent, see solutions

3. Dependent, see solutions

5. a. P(Jack/face card) = 0.333, b. P(heart/card a 3) = 0.25, c. P(red card/ace) = 0.50, d. not independent, see solutions, e. independent, see solutions

9. \(4.55 \times 10^{-4}\)

11. a. P(female) = 0.684, b. P(1st class) = 0.429, c. P(female/1st class) = 0.694, d. P(female and 1st class) = 0.298, e. P(female or 1st class) = 0.816, f. No, see solutions, g. Dependent, see solutions

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7.4.1: Conditional Probability (Exercises)

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• Page ID 135451

• Rupinder Sekhon and Roberta Bloom
• De Anza College

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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: