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3. The Relationship Between Independence and Conditional Probability

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eCourses /

Geometry /

Chapter 6

The Relationship Between Independence and Conditional Probability

Conditional probability and independence are two fundamental concepts in probability theory. The lesson delves into how these two are interconnected and how they differ. It uses sample space and tree diagrams as tools to explain these relationships. Understanding these concepts is crucial for anyone studying statistics, data science, or involved in decision-making processes that rely on probabilistic outcomes. The lesson serves as a comprehensive guide for both students and professionals who wish to grasp the nuances of conditional probability and independence.

Lesson Settings & Tools

	8 Theory slides
	6 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

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Sometimes, it is useful to identify the connection between different events by analyzing whether they are dependent or not. Another commonly used characteristic is a conditional probability. In this lesson, the conditional probability and its connection with independence will be discovered through the use of real-world examples.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Probability experiment
Probability of an event
Union
Intersection
Complement
Independent and dependent events

Challenge

Analyzing Connections Between Events

Vincenzo is thinking about taking a trip overseas. He is having a tough time deciding between France and Antarctica. In France he can visit the Eiffel Tower — he studies structural engineering — but Antartica has penguins and Vincenzo really loves penguins!

1. Vincenzo next to the Eiffel Tower. 2. Vincenzo in Antartica. 3. Vincenzo and a penguin

External credits: @pikisuperstar, @macrovector, @jcomp

Vincenzo pulls up a search engine and types "France vs. Antartica, I wanna see penguins." He finds an interesting survey that just so happens to contrast France and Antarctica. The survey even asks specifically about penguins!

Of $190$ people surveyed, $157$ people prefer going to France over Antarctica.
Of those $157,$ only five people saw penguins.
Of the people who traveled to Antarctica, only three of them did not see penguins.

The remaining results from the survey are organized in the following table.

Frequency table comparing people who go to France/Antarctica and see a penguin with those who do not

Consider the presented data to find the probabilities of the following scenarios.

a Vincenzo decides to travel to France. What is the probability that he will see a penguin? Round the probability to the nearest percent.

b Yesterday, Vincenzo saw a penguin! What is the probability that he was in France? Round the probability to the nearest percent.

c Vincenzo’s trip just ended. Sadly, he did not see a single penguin. What is the probability that he chose to go to Antarctica? Round the probability to the nearest percent.

Example

Relationship Between Dependent Events

Diego wants to pick two books at random from a pile of five books. Three of them are Geometry books, and the other two are History books. Below, the sample space of this situation is shown, where

G_{1},

G_{2},

G_{3}

represent the Geometry books, and

H_{1}

and

H_{2}

represent the History books.

Three Geometry books and Two History books

Imagine standing next to Diego and calculating the probabilities in real time with him. Let

H

be the event that the first book chosen is a History book, and

G

be the event that the second book chosen is a Geometry book. Now, consider the following scenarios.

a Find

P (H),

P (G),

and

P (H \cap G) .

Express the probabilities as fractions in their simplest form.

b Find

\frac{P ( H \cap G )}{P ( H )}

and

\frac{P ( H \cap G )}{P ( G )} .

Express the probabilities as fractions in their simplest form.

c If the first book Diego picked is a History book, what is the probability that the second book will be a Geometry book? Express the probability as a fraction in its simplest form.

d Diego chose both books before anyone else saw. If the second book Diego picked is a Geometry book, what is the probability that the first book he chose is a History book? Express the probability as a fraction in its simplest form.

e Is there any relationship between the probabilities found in Parts B, C, and D?

Answer

a Probabilities:

$P (H) = \frac{2}{5}$
$P (G) = \frac{3}{5}$
$P (H \cap G) = \frac{3}{1 0}$

\frac{P ( H \cap G )}{P ( H )} = \frac{3}{4}

and

\frac{P ( H \cap G )}{P ( G )} = \frac{1}{2} .

\frac{3}{4}

\frac{1}{2}

e Sample answer:

The probability that event $G$ happens given that event $H$ happened equals $\frac{P ( H \cap G )}{P ( H )} .$
The probability that event $H$ happens, given that event $G$ happened equals $\frac{P ( H \cap G )}{P ( G )} .$

Hint

a Every outcome that has a History book in the first position satisfies

H .

Similarly, every outcome that has a Geometry book in the second position satisfies

G .

Divide the number of favorable outcomes by the total number of outcomes.

b Divide the numbers found in Part A.

c The sample space of the given situation is not the original sample space. Start by finding the new sample space. Later, determine how many of the outcomes have a Geometry book in the second position.

d The sample space of the given situation is not the original sample space. Also, it is different from the one found in Part C. In this new sample space, count the outcomes that have a History book in the first position.

e Compare the probabilities found in Parts B, C, and D. Write a sentence stating the relationship found.

Solution

a By definition, the probability of an event is found by dividing the number of favorable outcomes by the total number of outcomes in the sample space.

P = \frac{Number of favorable outcomes}{Number of possible outcomes}

From the given diagram, the sample space contains

20

different outcomes. The required probabilities can be found one at a time.

To find $P (H),$ divide the number of outcomes in the sample space with a History book in the first position by $20 .$
To find $P (G),$ divide the number of outcomes in the sample space with a Geometry book in the second position by $20 .$
To find $P (H \cap G),$ divide the number of outcomes in the sample space with a History book in the first position and a Geometry book in the second position by $20 .$

The following diagram summarizes all the computations.

Notice that

P (H \cap G) \neq = P (H) \cdot P (G) .

That relationship implies that the given events are dependent.

b In this part, the probabilities found in Part A will be used.

P (H) P (G) P (H \cap G) = \frac{2}{5} = \frac{3}{5} = \frac{3}{1 0}

Start by finding the ratio of

P (H \cap G)

P (H) .

\frac{P ( H \cap G )}{P ( H )}

▼

Substitute values and simplify

SubstituteValues

Substitute values

\frac{\frac{3}{1 0}}{\frac{2}{5}}

DivFracByFracD

$\frac{a / b}{c / d} = \frac{a}{b} \cdot \frac{d}{c}$

\frac{3}{1 0} \cdot \frac{5}{2}

MultFrac

Multiply fractions

\frac{1 5}{2 0}

ReduceFrac

$\frac{a}{b} = \frac{a / 5}{b / 5}$

\frac{3}{4}

Next, find the ratio of

P (H \cap G)

P (G) .

\frac{P ( H \cap G )}{P ( G )}

▼

Substitute values and simplify

SubstituteValues

Substitute values

\frac{\frac{3}{1 0}}{\frac{3}{5}}

DivFracByFracD

$\frac{a / b}{c / d} = \frac{a}{b} \cdot \frac{d}{c}$

\frac{3}{1 0} \cdot \frac{5}{3}

MultFrac

Multiply fractions

\frac{1 5}{3 0}

ReduceFrac

$\frac{a}{b} = \frac{a / 1 5}{b / 1 5}$

\frac{1}{2}

c For this situation, it is known that the first book chosen is a History book. Knowing this fact reduces the number of possible outcomes because some of the outcomes are ignored in the new situation. Therefore, the new sample space contains fewer outcomes compared to the original one.

As seen, there are

8

outcomes in the new sample space and

6

of them have a Geometry book in the second position. Therefore, the probability that the second book is a Geometry book, given that the first book is a History book, can be found using the following process.

P (G given H) = \frac{6}{8} = \frac{3}{4}

Consequently, if the first book chosen by Diego is a History book, the probability that the second book is a Geometry book is

\frac{3}{4} .

d In this case, it is known that the second book chosen is a Geometry book. This reduces the number of possible outcomes from the original sample space. Therefore, the new sample space contains fewer outcomes and it is different from the one used in Part C.

As seen, there are

12

outcomes in the new sample space and

6

of them have a History book in the first position. Therefore, the probability that the first book is a History book, given that the second book is a Geometry book can be found as follows.

P (H given G) = \frac{6}{1 2} = \frac{1}{2}

Consequently, if the second book chosen by Diego is a Geometry book, the probability that the first book is a History book is

\frac{1}{2} .

e Start by writing the four probabilities found in Parts B, C, and D.

\frac{P ( H \cap G )}{P ( H )} \frac{P ( H \cap G )}{P ( G )} P (G given H) P (H given G) = \frac{3}{4} = \frac{1}{2} = \frac{3}{4} = \frac{1}{2}

Comparing the four probabilities, it can be seen that the first probability found in part B equals the probability found in part C.

The probability that the second book is a Geometry book given that the first book chosen is a History book equals $\frac{P ( H \cap G )}{P ( H )} .$

The previous statement can also be rewritten in terms of $H$ and $G$ as follows.

The probability that event $G$ happens given that event $H$ happened equals $\frac{P ( H \cap G )}{P ( H )} .$

Similarly, the second probability found in part B equals the probability found in part D. This leads to write the following relation.

The probability that the first book is a History book, given that the second book is a Geometry book equals $\frac{P ( H \cap G )}{P ( G )} .$

As before, the previous statement can be rewritten in terms of $H$ and $G .$

The probability that event $H$ happens, given that event $G$ happened equals $\frac{P ( H \cap G )}{P ( G )} .$

Example

Finding the Probability of Triggering an Alarm

At Troupial Airport, engineers are testing a prototype of a prohibited substance detector. If there is a forbidden item in a bag, an alarm is supposed to be triggered. To test the detector,

5000

bags will be checked and

7 %

of them will randomly contain forbidden items.

If a bag contains a forbidden item, there is a

98 %

chance that the alarm is triggered. On the other hand, if a bag does not have a forbidden item, there is a

4 %

chance that it triggers the alarm. Calculate the probabilities of the events described below. Write the probabilities as a percent rounded to two decimal places, when necessary.

a Find the probability that a bag triggers the alarm.

b Find the probability that a bag triggers the alarm and contains a forbidden item. Also, find the probability that a bag does not trigger the alarm and contains a forbidden item.

c If Mark's bag triggered the alarm, what is the probability that his bag contains a forbidden item?

d If Izabella's bag did not trigger the alarm, what is the probability that her bag contains a forbidden item?

e Based on Parts C and D, what can be said about Mark and Izabella situations?

Answer

P (Alarm) = 10.58 %

P (Alarm and Forbidden) = 6.86 %

and

P (No Alarm and Forbidden) = 0.14 %

c The probability that Mark's bag contains a forbidden item given that it triggered the alarm is about

64.84 % .

d The probability that Izabella's bag contains a forbidden item given that it did not trigger the alarm is about

0.16 % .

e Example Solution:

There is a $64.84 %$ chance that Mark's bag contains a forbidden item. However, this probability is not big enough to ensure Mark has a forbidden item. Therefore, it is doubtful — but possible — that Mark contains a forbidden item in his bag.
There is a $0.16 %$ chance that Izabella's bag contains a forbidden item. Since this probability is very small — less than $1 %$ — it is almost certain that Izabella does not have forbidden items in her bag — but still possible.

Hint

a Make a tree diagram and write the number of bags corresponding to every node. The probability of triggering the alarm is the total number of bags that triggered the alarm divided by the total amount of bags checked.

b Determine how many bags contain a forbidden item and trigger the alarm. Divide this number by the total number of bags checked. Similar reasoning can be used for the case where the bags do not trigger the alarm.

c Considering the diagram drawn in Part A, what is the total number of bags that trigger the alarm? Of all these bags, how many actually had forbidden items?

d Of all the bags that did not trigger the alarm, how many had forbidden items? Divide this last number by the total number of bags that did not trigger the alarm.

e If the probability of an event is closer to

50 %

than

100 %,

then it is doubtful that the event actually happened. If the probability of an event is close to

0 %,

it is almost certain that the event did not happen.

Solution

a To find the probability that a bag triggers the alarm, a tree diagram will be drawn. To do so, first notice that the situation can be divided into three different stages.

All of the previous stages and events can be illustrated using a tree diagram.

Recall that there are

5000

bags, and

7 %

of them contain forbidden items. The product of these numbers will give the number of the bags that contain forbidden items. The rest of the bags do not contain forbidden items.

Forbidden : Not Forbidden : 5000 \cdot 7 % = 350 5000 - 350 = 4650

Additionally, the following two details about the bags are known.

If a bag contains forbidden items, there is a $98 %$ chance that it triggers the alarm.
If a bag does not have forbidden items, there is a $4 %$ chance that it triggers the alarm.

Considering these details, it can be concluded that $2 %$ of the bags containing forbidden items could trigger the alarm and $96 %$ of the bags that do not have forbidden items could not trigger the alarm.

Now, using the percentages in the branches, the number of bags for each event can be found.

Forbidden and Alarm	$350 \cdot 98 % = 343$
Forbidden and No Alarm	$350 \cdot 2 % = 7$
Not Forbidden and Alarm	$4650 \cdot 4 % = 186$
Not Forbidden and No Alarm	$4650 \cdot 96 % = 4464$

Finally, all the information can be shown on the tree diagram.

To find the probability that a bag triggers the alarm, start by finding how many bags triggered the alarm. To do so, add the two corresponding numbers in the tree diagram.

Bags that triggered the alarm : 343 + 186 = 529

The probability that a randomly picked bag triggers the alarm

P (Alarm)

is obtained dividing

529

by the total number of bags.

P (Alarm) = \frac{5 2 9}{5 0 0 0}

CalcQuot

Calculate quotient

P (Alarm) = 0.1058

WritePercent

Convert to percent

P (Alarm) = 10.58 %

b Pulling information from the tree diagram, it is seen that the number of bags that triggered the alarm and had forbidden items is

343

and the number of bags that contained forbidden items and did not trigger the alarm is

7 .

Since the total number of bags is

5000,

the ratio of the number of the favorable outcomes to the total number will give the desired probability.

Probabilities of the Events
$P (A l a r m and F o r b i d d e n)$	$\frac{3 4 3}{5 0 0 0} = 6.86 %$
$P (N o A l a r m and F o r b i d d e n)$	$\frac{7}{5 0 0 0} = 0.14 %$

Take note that the sum of the probabilities is equal to $7 %,$ which is the percentage of the bags that contain forbidden items.

c To determine the probability that a bag contains forbidden items — given that it triggered the alarm — divide the total number of bags that both triggered the alarm and had forbidden items by the total number of bags that triggered the alarm.

P = \frac{A l a r m and F o r b i d d e n}{Alarm}

As presented in the tree diagram, the number of bags that trigger the alarm is equal to the sum of the numbers under the word Alarm.

Therefore, there are

529

bags that trigger the alarm. Since Mark's bag triggered the alarm, his bag is one of those

529 .

Out of this group of bags,

343

had forbidden items.

P = \frac{3 4 3}{5 2 9}

▼

Evaluate right-hand side

CalcQuot

Calculate quotient

P = 0.648393 \dots

RoundDec

Round to $4$ decimal place(s)

P \approx 0.6484

WritePercent

Convert to percent

P \approx 64.84 %

Therefore, there is about a

64.84 %

chance that Mark's bag contains forbidden items given that his bag triggered the alarm.

d To determine the probability that a bag contains forbidden items given that it did not trigger the alarm, divide the total number of bags that did not trigger the alarm and had forbidden items by the total number of bags that did not trigger the alarm.

P = \frac{N o A l a r m and F o r b i d d e n}{No Alarm}

First, find the number of bags did not trigger the alarm.

Since Izabella's bag did not trigger the alarm, her bag is one of those

4471 .

Out of these bags,

7

had forbidden items.

P = \frac{7}{4 4 7 1}

▼

Evaluate right-hand side

CalcQuot

Calculate quotient

P = 0.001565 \dots

RoundDec

Round to $4$ decimal place(s)

P \approx 0.0016

WritePercent

Convert to percent

P \approx 0.16 %

Therefore, there is about a

0.16 %

chance that Izabella's bag contains forbidden items given that her bag did not trigger the alarm.

e If the probability of an event is closer to

50 %

than

100 %,

then it is doubtful that the event would actually happen. With that in mind, consider the probability found in Part C.

There is about a $64.84 %$ chance that Mark's bag contains a forbidden item.

This probability is not close enough to $100 %$ to ensure that Mark's bag contains a forbidden item. Therefore, it is doubtful — but possible — that Mark's bag contains a forbidden item. Next, recall the answer found in Part D.

There is about a $0.16 %$ chance that Izabella's bag contains a forbidden item.

Since this probability is very small — less than $1 %$ — it is almost certain that Izabella does not have forbidden items in her bag — but still possible.

In this example, as well as in the previous one, dependent events and how the occurrence of one event affects the occurrence of the other event were analyzed through counting methods like tables and tree diagrams. Below, a formula for finding the probability of similar situations is shown.

Discussion

Conditional Probability

Conditional probability is the measure of the likelihood of an event $B$ occurring, given that event $A$ has occurred previously. The probability of $B$ given $A$ is written as $P (B ∣ A) .$ It can be calculated by dividing the probability of the intersection of $A$ and $B$ by the probability of $A .$

$P (B ∣ A) = \frac{P ( A and B )}{P ( A )}, where P (A) \neq = 0$

It is worth noting that usually

P (B ∣ A)

and

P (A ∣ B)

are not equal, meaning that conditional probability is not reversible. For example, let

A

be the event of a prime number and

B

be the event of an odd number. The probability that a prime number is odd is almost

1,

but the reverse, an odd number being prime, is much smaller.

P (B ∣ A) \neq = P (A ∣ B)

Why

Intuition Behind the Formula

The intuition behind the formula can be visualized by using Venn Diagrams. Consider a sample space $S$ and the events $A$ and $B$ such that $P (A) \neq = 0 .$

Assuming that event $A$ has occurred, the sample space is reduced to $A .$

This means that the probability that event $B$ can happen is reduced to the outcomes in the intersection of $A$ and $B,$ that is, to those outcomes in $A \cap B .$

The possible outcomes are given by $P (A)$ and the favorable outcomes by $P (A \cap B) .$ Therefore, the conditional probability formula can be obtained using the probability formula.

$P (B ∣ A) = \frac{P ( A and B )}{P ( A )}$

Now that the formula for finding a conditional probability is known, it can be put to use to solve daily life situations.

Example

Probability of Getting the Correct Size and Color

Diego's generous father has finished doing laundry and put Diego's T-shirts along with those of his big brother into the same ol’ basket. There are orange, blue, and red T-shirts in the basket, of which four are S-sized and eight are M-sized.

S-sized shirts: 1 Orange, 1 Blue, 2 Red; M-sized shirts: 5 Orange, 2 Blue, 1 Red

Diego is planning to go out with his friends. After taking a shower, he randomly picks a T-shirt from the same ol' laundry basket. Consider the following events.

Event S : Event O : Event B : Diego picks an S - sized T - shirt . Diego picks an orange T - shirt . Diego picks a blue T - shirt .

a Find and compare

P (S ∣ O)

and

P (O ∣ S) .

Write the answers as fractions in their simplest form.

b Find and compare

P (S ∣ B)

and

P (B ∣ S) .

Write the answers as fractions in their simplest form.

Hint

a Start by finding

P (O),

P (S),

and

P (S and O) .

Then, apply the formula for finding the conditional probability.

b Find

P (B)

and

P (S and B) .

From Part A,

P (S)

is already known. Apply the formula for finding conditional probability again.

Solution

a To start, remember the formula for finding the conditional probability that event

B

happens given that event

A

has occurred.

P (B ∣ A) = \frac{P ( A and B )}{P ( A )}

Applying this formula, the required probabilities can be rewritten as follows.

P (S ∣ O) P (O ∣ S) = \frac{P ( S and O )}{P ( O )} = \frac{P ( S and O )}{P ( S )}

Consequently,

P (S and O),

P (O),

and

P (S)

need to be found. To do so, start by writing the formula to find the probability of an event.

P = \frac{Number of favorable outcomes}{Number of possible outcomes}

Notice that

P (S and O)

represents the probability of picking an S-size

orange

T-shirt. From the diagram, there are

12

T-shirts in the basket, and there is only one S-size

orange

T-shirt.

P (S and O) = \frac{1}{1 2} ✓

To find

P (O),

count how many of the

12

T-shirts are

orange .

From the diagram, there are only

5 .

P (O) = \frac{5}{1 2} ✓

To determine

P (S),

count how many of the

12

T-shirts are S-sized. From the diagram, there are only

4 .

P (S) = \frac{4}{1 2} = \frac{1}{3} ✓

Finding Conditional Probabilities

To find

P (S ∣ O),

divide

P (S and O) = \frac{1}{1 2}

P (O) = \frac{5}{1 2} .

P (S ∣ O) = \frac{P ( S and O )}{P ( O )}

▼

Substitute values and simplify

SubstituteII

$P (S and O) = \frac{1}{1 2}$ , $P (O) = \frac{5}{1 2}$

P (S ∣ O) = \frac{\frac{1}{1 2}}{\frac{5}{1 2}}

DivFracByFracD

$\frac{a / b}{c / d} = \frac{a}{b} \cdot \frac{d}{c}$

P (S ∣ O) = \frac{1}{1 2} \cdot \frac{1 2}{5}

MultFrac

Multiply fractions

P (S ∣ O) = \frac{1 2}{6 0}

ReduceFrac

$\frac{a}{b} = \frac{a / 1 2}{b / 1 2}$

P (S ∣ O) = \frac{1}{5}

Therefore, the probability that Diego picks an S-size T-shirt given that he picked a orange one is

\frac{1}{5},

20 % .

Finally, find

P (O ∣ S) .

P (O ∣ S) = \frac{P ( S and O )}{P ( S )}

▼

Substitute values and simplify

SubstituteII

$P (S and O) = \frac{1}{1 2}$ , $P (S) = \frac{1}{3}$

P (O ∣ S) = \frac{\frac{1}{1 2}}{\frac{1}{3}}

DivFracByFracD

$\frac{a / b}{c / d} = \frac{a}{b} \cdot \frac{d}{c}$

P (O ∣ S) = \frac{1}{1 2} \cdot \frac{3}{1}

MultFrac

Multiply fractions

P (O ∣ S) = \frac{3}{1 2}

ReduceFrac

$\frac{a}{b} = \frac{a / 3}{b / 3}$

P (O ∣ S) = \frac{1}{4}

Consequently, the probability that Diego picks a

orange

T-shirt given that he picked an S-size one is

\frac{1}{4},

25 % .

b Similar to the previous part, start by applying the conditional probability formula to rewrite the given expressions.

P (S ∣ B) P (B ∣ S) = \frac{P ( S and B )}{P ( B )} = \frac{P ( S and B )}{P ( S )}

From Part A, it is known that

P (S)

equals

\frac{1}{3} .

Thus, only

P (S and B)

and

P (B)

are missing. Note that

P (S and B)

is the probability of picking an S-size

b l u e

T-shirt. From the

12

T-shirts, only two are

b l u e

and S-sized.

P (S and B) = \frac{2}{1 2} = \frac{1}{6} ✓

To determine

P (B),

count how many of the

12

T-shirts are

b l u e .

From the diagram, there are

4 .

P (B) = \frac{4}{1 2} = \frac{1}{3} ✓

Finding Conditional Probabilities

Substituting the found probabilities into the initial equations, the required probabilities will be obtained. Start by finding

P (S ∣ B) .

P (S ∣ B) = \frac{P ( S and B )}{P ( B )}

▼

Substitute values and simplify

SubstituteII

$P (S and B) = \frac{1}{6}$ , $P (B) = \frac{1}{3}$

P (S ∣ B) = \frac{\frac{1}{6}}{\frac{1}{3}}

DivFracByFracD

$\frac{a / b}{c / d} = \frac{a}{b} \cdot \frac{d}{c}$

P (S ∣ B) = \frac{1}{6} \cdot \frac{3}{1}

MultFrac

Multiply fractions

P (S ∣ B) = \frac{3}{6}

ReduceFrac

$\frac{a}{b} = \frac{a / 3}{b / 3}$

P (S ∣ B) = \frac{1}{2}

Therefore, the probability that Diego picks an S-size T-shirt given that he picked a blue one is

\frac{1}{2},

50 % .

Finally, find

P (B ∣ S) .

P (B ∣ S) = \frac{P ( S and B )}{P ( S )}

▼

Substitute values and simplify

SubstituteII

$P (S and B) = \frac{1}{6}$ , $P (S) = \frac{1}{3}$

P (B ∣ S) = \frac{\frac{1}{6}}{\frac{1}{3}}

DivFracByFracD

$\frac{a / b}{c / d} = \frac{a}{b} \cdot \frac{d}{c}$

P (B ∣ S) = \frac{1}{6} \cdot \frac{3}{1}

MultFrac

Multiply fractions

P (B ∣ S) = \frac{3}{6}

ReduceFrac

$\frac{a}{b} = \frac{a / 3}{b / 3}$

P (B ∣ S) = \frac{1}{2}

Consequently, the probability that Diego picks a

b l u e

T-shirt given that he picked an S-size one is

\frac{1}{2},

50 % .

Time go to go out strutting wearing one of the shirts from the same old laundry basket — hopefully it is not his brothers!

Pop Quiz

Applying the Conditional Probability Formula

Find the required conditional probability and round it to two decimal places.

Random Probabilities of A,B, and 'A and B'

Closure

Destinations, Penguins, and Probabilities

Now that it is known how to compute conditional probabilities, Vincenzo's situation can be better investigated.

1. Tearrik next to the Eiffel Tower. 2. Tearrik in Antartica. 3. Tearrik and a penguin. 4. Frequency table comparing people who go to France/Antarctica and see a penguin with those who do not

External credits: @pikisuperstar, @macrovector, @jcomp

a Vincenzo decides to travel to France. What is the probability he will see a penguin? Round the probability to the nearest percent.

b Yesterday, Vincenzo saw a penguin! What is the probability that he was in France? Round the probability to the nearest percent.

c Vincenzo’s trip just ended. Sadly, he did not see a single penguin. What is the probability that he chose to go to Antarctica? Round the probability to the nearest percent.

Hint

a Use the formula of conditional probability. Gathering the data from the table, a total of

190

people participated in the survey. Of that total

5,

traveled to France and saw penguins.

b A total of

35

people saw penguins. How many of these were in France?

c Of the

190

people,

155

did not see penguins and

3

of them traveled to Antarctica.

Solution

a By the definition of conditional probability, the probability that Vincenzo saw a penguin given that he traveled to France can be expressed in the following manner.

To find the corresponding probabilities, take a look at the table.

A total

190

people participated in the survey and

5

of them traveled to France and saw a penguin.

P (Penguin and France) = \frac{5}{1 9 0} = \frac{1}{3 8}

Additionally,

157

people traveled to France.

P (France) = \frac{1 5 7}{1 9 0}

Next, substitute these two probabilities into the conditional probability formula.

P (Penguin ∣ France) = \frac{P ( Penguin and France )}{P ( France )}

▼

Substitute values and simplify

SubstituteII

$P (Penguin and France) = \frac{1}{3 8}$ , $P (France) = \frac{1 5 7}{1 9 0}$

P (Penguin ∣ France) = \frac{\frac{1}{3 8}}{\frac{1 5 7}{1 9 0}}

DivFracByFracD

$\frac{a / b}{c / d} = \frac{a}{b} \cdot \frac{d}{c}$

P (Penguin ∣ France) = \frac{1}{3 8} \cdot \frac{1 9 0}{1 5 7}

MultFrac

Multiply fractions

P (Penguin ∣ France) = \frac{1 9 0}{5 9 6 6}

CalcQuot

Calculate quotient

P (Penguin ∣ France) = 0.031847 \dots

WritePercent

Convert to percent

P (Penguin ∣ France) = 3.1847 \dots %

RoundInt

Round to nearest integer

P (Penguin ∣ France) \approx 3 %

Consequently, Vincenzo has about a

3 %

chance of seeing penguins, given that he traveled to France.

b In this case, the situation is opposite to that presented in Part A. Now, it is known that Vincenzo saw penguins, and it is asked to find the probability that Vincenzo is in France. As before, start by applying the conditional probability formula.

From Part A, the numerator is equal to

\frac{1}{3 8} .

To determine the probability of seeing penguins, determine how many of the

190

people surveyed actually saw penguins. According to the table,

35

people answered that they did.

P (Penguins) = \frac{3 5}{1 9 0} = \frac{7}{3 8}

The next step is to substitute the two probabilities found into the conditional probability formula.

P (France ∣ Penguin) = \frac{P ( Penguin and France )}{P ( Penguins )}

▼

Substitute values and simplify

SubstituteII

$P (Penguin and France) = \frac{1}{3 8}$ , $P (Penguins) = \frac{7}{3 8}$

P (France ∣ Penguin) = \frac{\frac{1}{3 8}}{\frac{7}{3 8}}

DivFracByFracD

$\frac{a / b}{c / d} = \frac{a}{b} \cdot \frac{d}{c}$

P (France ∣ Penguin) = \frac{1}{3 8} \cdot \frac{3 8}{7}

MultFrac

Multiply fractions

P (France ∣ Penguin) = \frac{3 8}{2 6 6}

CalcQuot

Calculate quotient

P (France ∣ Penguin) = 0.142857 \dots

WritePercent

Convert to percent

P (France ∣ Penguin) = 14.2857 \dots %

RoundInt

Round to nearest integer

P (France ∣ Penguin) \approx 14 %

In conclusion, there is about a

14 %

chance that Vincenzo was in France, given that he saw penguins.

c Once more, start by applying the conditional probability formula.

From the second row and second column of the table,

3

of the

190

people traveled to Antarctica and did not see penguins.

P (Antarctica and No Penguin) = \frac{3}{1 9 0}

Seen in the third row of the table,

155

people did not see penguins. Knowing this, the probability that a person picked at random did not see penguins can be computed.

P (No Penguin) = \frac{1 5 5}{1 9 0} = \frac{3 1}{3 8}

Finally, substitute these values into the conditional probability formula.

P (Antarctica ∣ No Penguin) = \frac{P ( Antarctica and No Penguin )}{P ( No Penguin )}

▼

Substitute values and simplify

SubstituteII

$P (Antarctica and No Penguin) = \frac{3}{1 9 0}$ , $P (No Penguin) = \frac{3 1}{3 8}$

P (Antarctica ∣ No Penguin) = \frac{\frac{3}{1 9 0}}{\frac{3 1}{3 8}}

DivFracByFracD

$\frac{a / b}{c / d} = \frac{a}{b} \cdot \frac{d}{c}$

P (Antarctica ∣ No Penguin) = \frac{3}{1 9 0} \cdot \frac{3 8}{3 1}

MultFrac

Multiply fractions

P (Antarctica ∣ No Penguin) = \frac{1 1 4}{5 8 9 0}

CalcQuot

Calculate quotient

P (Antarctica ∣ No Penguin) = 0.019354 \dots

WritePercent

Convert to percent

P (Antarctica ∣ No Penguin) = 19.354 \dots %

RoundInt

Round to nearest integer

P (Antarctica ∣ No Penguin) \approx 2 %

Consequently, there is about a

2 %

chance that Vincenzo chose to go to Antarctica, given that he did not see penguins.

Level 1

Level 2

Level 3

The Relationship Between Independence and Conditional Probability

Exercise 1.1

A jar contains blue and red marbles.

a jar of red and blue marbles of different sizes

Calculate

P (large ∣ red) .

Answer with a fraction in its simplest form.

Calculate

P (blue ∣ large) .

Answer with a fraction in its simplest form.

Which of the conditional probabilities is greater,

P (red ∣ small)

P (small ∣ red) ?

To calculate P(large|red), we will use the following formula. P( A| B)=P( A and B)/P( B) If we apply this formula to our situation, we get the following equation. P(large|red)=P( large and red)/P(red) The probability of drawing a red marble is the ratio of the number of red marbles to the total number of marbles. Examining the jar, we count 6 red marbles and 10 marbles in total. Now we can determine P(red).

Let's also calculate the probability of drawing a large marble that is also red. We have 2 large red marbles. With this information, we can calculate P(large and red).

Now we have everything we need to calculate P(large|red).

As in the previous part, we must determine two probabilities in order to calculate the given conditional probability. The jar contains 4 blue marbles. We also see that 2 of the blue marbles are large. P(blue and large)&=2/10=1/5 [0.7em] P(large)&=4/10=2/5 Now we can calculate the given conditional probability.

Here we have two conditional probabilities we must determine. P(red|small)&=P(red and small)/P(small) [2em] P(small|red)&=P(small and red)/P(red) Note that the numerators of both fractions on the right-hand side of the equations are equivalent. We already know the probability of a marble being red is 35. We have 6 small marbles, of which 4 are red. With this information, we can determine the remaining probabilities. P(red and small)&=4/10=2/5 [1em] P(small)&=6/10=3/5 Now we have all the information we need to determine both probabilities.

Let's next determine the second probability.

As we can see, the probabilities are the same. Note that this did not have to be the case though. There is nothing that states that P(B|A) must be equal to P(A|B).

Jordan is employed as a mechanic. Every month she does yearly inspections of cars. The following table shows the result of a month's work.

two way table survey showing inspections of cars

Use the information to calculate the following conditional probabilities. Answer with a fraction in its simplest form.

P (pass ∣ defective)

P (fail ∣ non - defective)

The expression P(pass|defective) is the probability of a car passing the inspection given that it was defective. To determine this, we must use the following formula. P( A| B)=P( A and B)/P( B) If we apply this formula to our given situation, we get the following equation.

To determine the probabilities in the numerator and denominator of the fraction, we have to divide the number of cars that fits the given descriptions by the total number of cars. To do this, let's first calculate the total number of cars inspected.

By adding the all the numbers in the table we can determine the total number of cars that were inspected during the month. 2+ 40+ 450+ 10=502 Now we can find each probability.

Let's next determine the probability of a car getting a pass and being defective.

Now we can determine the conditional probability.

To calculate this conditional probability, we must divide the probability of a car failing the inspection and being non-defective by the probability that a car is non-defective. 450+ 10=460 From the table, we see that 460 cars were non-defective.

Let's also calculate the probability of a car both failing the inspection and being non-defective. From the table, we see that 10 cars failed the inspection while being non-defective.

Now we can determine the given conditional probability.

Consider the following sample space.

x = {5, 6, 7, 8, 9, 10, 11, 12, 13, 14}

Find the following probabilities. Answer with a fraction in its simplest form.

P (x < 10 ∣ x < 13)

P (x \geq 7 ∣ x < 13)

The expression P(x<10|x<13) is the probability of an observation in the sample space being less than 10, given that the observation is less than 13. To calculate this, we will use the following formula. P( A| B)=P( A and B)/P( B) If we apply this formula to the given situation, we get the following equation. P( x < 10| x < 13) = P( x < 10 and x < 13)/P( x < 13) We need to determine the probabilities in the numerator and denominator of the fraction. We have a total of ten observations and eight of them are below 13.

Let's also calculate the second probability.

Now we can determine the given conditional probability.

This time, we need to determine the probability of an observation being less than thirteen as well as the probability of an observation being greater than or equal to seven and less than thirteen. From the previous section we know the probability of a number being less than thirteen. P(x<13)=4/5 Examining the diagram, we see that six observations are greater than or equal to seven and also less than thirteen. With this information, we can determine the probability.

Now we can determine the given conditional probability.

The Relationship Between Independence and Conditional Probability

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