Diving into Applications of Conditional Probability: Venn Diagram, Table, and Tree Diagram Insights

In this lesson, a variety of conditional situations will be modeled by using tree diagrams, frequency tables, and Venn diagrams.

Catch-Up and Review

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

Probability
Tree Diagram
Venn Diagram
Frequency Table
Uniform Probability Model
Conditional Probability

Challenge

Detecting if an Email Is Spam

Spam filters determine whether an email is spam by checking it for some words that appear more frequently in spam emails. The following set of information is known.

$50 %$ of emails are spam.
$15 %$ of spam emails contain the word free.
$1 %$ of non-spam emails contain the word free.

Jordan gets an email with the word free in it.

a Draw a tree diagram to visualize the situation.

b What is the probability that the email is spam?

Discussion

Reviewing the Intuition Behind the Conditional Probability Formula

Recall the formula for the conditional probability.

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

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 $P (A) \neq = 0 .$

Venn diagram for sample space S with events A and B.

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

Events A and B, given that A has occurred

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

A Venn diagram that highlights the intersection of events A and B within the sample space S.

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.

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

Example

Using a Venn Diagram to Find Conditional Probability

Dominika and her friends, $10$ people in total, want to play basketball. They decide to form two teams randomly. To do so, each draws a card from a stack of $10$ cards numbered from $1$ to $10 .$

People who draw odd numbers will be on Team Red.
People who draw even numbers will be on Team Blue.

a Dominika excitedly draws the first card. What is the probability that she draws the number

1 ?

Write the probability as a fraction in its simplest form.

b If Dominika is on Team Red, what is the probability that she drew the number

1 ?

Write the probability as a fraction in its simplest form.

Hint

a Determine the sample space.

b Start by drawing a Venn diagram to represent the situation. Use the diagram to determine the sample space of the situation.

Solution

a Since Dominika is the first one to draw a card, there are

10

options for her. Therefore, the sample space for this case consists of numbers from

1

10 .

S = {1, 2, 3, \dots, 10}

The favorable outcome is to draw the number

1 .

Using the Probability Formula, the probability of drawing

1

is the ratio of the number of favorable outcomes

1

to the total number of outcomes

10 .

P (drawing 1) = \frac{1}{1 0}

b The given situation will be represented by using a Venn diagram. The following two events will be examined.

A : B : drawing an odd number drawing the number 1

The rest of the possible outcomes will be shown outside of the events

A

and

B .

Since Dominika is on Team Red, the sample space is reduced to the outcomes in $A .$

There are $five$ odd numbers in the new sample space and only $one$ favorable outcome.

Therefore, using the Probability Formula, the probability that Dominika drew the number

1,

given that the number drawn is odd, is found as follows.

P (B ∣ A) = \frac{1}{5}

Pop Quiz

Practice Finding Conditional Probability

The applet shows the probabilities of two events in a Venn diagram. Calculate the conditional probabilities. If necessary, round the answer to two decimal places.

Probabilities of events A and B shown in Venn diagram

It is practical to model many real-life situations with a uniform probability model, even though they may not actually be uniform after close inspection. Examine the next example and interpret the results.

Example

Using a Table to Find Conditional Probabilities

After reading an article about the famous wreck of the Titanic, Paulina concluded that the rescue procedures favored the wealthier first-class passengers. She then finds some data on the survival of the Titanic passengers.

	Survived	Did Not Survive	Total
First Class Passengers	$201$	$123$	$324$
Second Class Passengers	$118$	$166$	$284$
Third Class Passengers	$181$	$528$	$709$
Total	$500$	$817$	$1317$

Use this data to investigate the probabilities of surviving the wreck of the Titanic.

a Determine if the events are independent or not. Justify your answer using appropriate probability calculations.

	$A :$ Passenger Survived
$B :$ First Class Passenger
$C :$ Second Class Passenger
$D :$ Third Class Passenger

b Did all passengers aboard the Titanic have the same probability of surviving? Justify your answer.

Answer

a Table:

	$A :$ Passenger Survived
$B :$ First Class Passenger	Dependent, $P (A ∣ B) \neq = P (A)$
$C :$ Second Class Passenger	Dependent, $P (A ∣ C) \neq = P (A)$
$D :$ Third Class Passenger	Dependent, $P (A ∣ D) \neq = P (A)$

b No, see solution.

Hint

a Use the fact that two events

A

and

B

are independent if

P (A ∣ B) = P (A) .

b Compare the conditional probabilities

P (A ∣ B),

P (A ∣ C),

and

P (A ∣ D) .

Solution

a If two events are independent, the occurrence of one of the events does not affect the occurrence of the other. In other words, the probability that

A

occurs given that

B

has already occurred is the same as the probability that

A

occurs.

P (A ∣ B) = P (A)

Therefore, the following probabilities need to be calculated first.

P (A), P (A ∣ B), P (A ∣ C), and P (A ∣ D)

To do so, the Probability Formula will be used.

P = \frac{Number of favorable outcomes}{Total number of outcomes}

Now, examine the given table to find the probabilities of each event. Adding the event labels to the table can make finding the probabilities a little easier. Recall that when finding a conditional probability, the sample space is reduced.

Having found the probabilities, note that none of the conditional probabilities is equal to

P (A) .

	Fraction	Decimal
$P (A)$	$\frac{5 0 0}{1 3 1 7}$	$\approx 0.380$
$P (A ∣ B)$	$\frac{2 0 1}{3 2 4}$	$\approx 0.620$
$P (A ∣ C)$	$\frac{1 1 8}{2 8 4}$	$\approx 0.415$
$P (A ∣ D)$	$\frac{1 8 1}{7 0 9}$	$\approx 0.255$

Therefore, events $B,$ $C,$ and $D$ each have an effect on event $A,$ meaning that $A$ is a dependent event. In simpler terms, a passenger's chance of surviving depended on what class they were traveling in.

	$A :$ Passenger Survived
$B :$ First Class Passenger	Dependent, $P (A ∣ B) \neq = P (A)$
$C :$ Second Class Passenger	Dependent, $P (A ∣ C) \neq = P (A)$
$D :$ Third Class Passenger	Dependent, $P (A ∣ D) \neq = P (A)$

b In the previous part, the conditional probabilities

P (A ∣ B),

P (A ∣ C),

and

P (A ∣ D)

were found.

P (A ∣ B) \approx 0.620 P (A ∣ C) \approx 0.415 P (A ∣ D) \approx 0.255

Comparing these probabilities, it can be concluded that not all passengers aboard the Titanic had the same chance of surviving.

P (A ∣ B) > P (A ∣ C) > P (A ∣ D) 0.620 > 0.415 > 0.255

The first class passengers had the greatest chance of being rescued.

It would be appropriate to assume that each passenger had the same chance of survival, as the uniform probability model suggests. However, the data shows that not all passengers had the same chance of survival.

Pop Quiz

Practice Finding Conditional Probability

The applet shows the frequency of each event in a table. Calculate the conditional probability asked in the applet. If necessary, round the answer to two decimal places.

Example

Using a Tree Diagram to Find Conditional Probabilities

Tadeo searches the Internet to check how effective Drug A is compared to Drug B. He finds a research paper about the drugs that gives the following information.

One third of the participants received Drug A, one third received Drug B, and one third received a placebo.
$7 %$ of participants who received Drug A, $11 %$ of participants who Drug B, and $5 %$ of participants who received the placebo reported recovery from their condition.

Help Tadeo answer the following questions.

a Draw a tree diagram to represent the situation.

b What is the probability that a participant did not recover if they received Drug B?

c Suppose that the number of participants is

6000 .

What is the number of participants that received Drug A and did not recover?

Answer

a Example Tree Diagram:

P (N R ∣ B) = 0.67

1580

Hint

a Start by considering the first point of the information Tadeo found. How many outcomes are there initially?

b Determine if it is a conditional probability or not. Use the tree diagram to find it.

c Start by finding the probability of a participant receiving Drug A and not recovering.

Solution

a Consider the first point of the information found by Tadeo.

One third of the participants received Drug A, one third recieved Drug B, and one third received a placebo.

Since there are three possible outcomes, start by drawing three branches and labeling them.

A : B : C : Receiving Drug A Receiving Drug B Receiving the placebo

Additionally, since the drugs were distributed among participants evenly, each branch has the same probability,

\frac{1}{3} .

Each of these outcomes will have two further outcomes — recovered $R$ or not recovered $N R .$ Therefore, two more branches will be drawn for each case.

Each branch should have a probability value on it. Notice that these probabilities are conditional probabilities.

Since there are only two possible outcomes after a certain medicine is received, these outcomes are complements of each other. Therefore, by the Complement Rule, the following equations can be written.

P (N R ∣ A) = 1 - P (R ∣ A) P (N R ∣ B) = 1 - P (R ∣ B) P (N R ∣ C) = 1 - P (R ∣ C)

Next, these conditional probabilities can be calculated using the information from the second point.

$7 %$ of participants who received Drug A, $11 %$ of participants who Drug B, and $5 %$ of participants who received the placebo reported recovery from their condition.

Knowing that

one

third

of the participants received Drug A and

7 %,

0.07,

of the participants who received Drug A recovered,

P (R ∣ A)

can be calculated using the Conditional Probability Formula.

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

SubstituteII

$P (A) = \frac{1}{3}$ , $P (R and A) = 0.07$

P (R ∣ A) = \frac{0 . 0 7}{\frac{1}{3}}

▼

Simplify right-hand side

DivByFracD

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

P (R ∣ A) = \frac{0 . 0 7 \cdot 3}{1}

Multiply

P (R ∣ A) = \frac{0 . 2 1}{1}

DivByOne

$\frac{a}{1} = a$

P (R ∣ A) = 0.21

When a participant receives Drug A,

0.21

of them recover, which means that

0.79

do not recover. The other conditional probabilities can be found in a similar fashion. The percentages will be written as decimals when calculating each probability.

11 % = 0.11 5 % = 0.05

The other conditional probabilities can be found in the table.

	Conditional Probability Formula	Substitute	Evaluate	Probability of the Complement
$P (R ∣ A)$	$\frac{P ( R and A )}{P ( A )}$	$\frac{0 . 0 7}{\frac{1}{3}}$	$0.21$	$1 - 0.21 = 0.79$
$P (R ∣ B)$	$\frac{P ( R and B )}{P ( B )}$	$\frac{0 . 1 1}{\frac{1}{3}}$	$0.33$	$1 - 0.33 = 0.67$
$P (R ∣ C)$	$\frac{P ( R and C )}{P ( C )}$	$\frac{0 . 0 5}{\frac{1}{3}}$	$0.15$	$1 - 0.15 = 0.85$

Finally, the tree diagram can be completed.

b A participant's well-being varies depending on the drug they received. The probability that a participant did not recover if they received Drug B is a conditional probability because knowing that a participant received Drug B changes the probability of recovery.

P (N R ∣ B)

Recall that each final branch of the tree diagram represents a conditional probability. The branch between

B

and

N R

will represent the probability of a participant receiving Drug B and not recovering.

Therefore, the conditional probability that a participant did not recover if they received Drug B is $0.67 .$

c Consider the path through

A

N R .

The highlighted path represents

P (A and N R),

the probability that a participant received Drug A and did not recover. This probability is equal to the product of the probabilities on the branches.

P (A and N R) = \frac{1}{3} \cdot 0.79

The number of study participants who received Drug A and did not recover can be found by multiplying this probability by the total number of participants,

6000

people.

\frac{1}{3} \cdot 0.79 \cdot 6000

▼

Simplify

Multiply

\frac{1}{3} \cdot 4740

MoveRightFacToNumOne

$\frac{1}{b} \cdot a = \frac{a}{b}$

\frac{4 7 4 0}{3}

CalcQuot

Calculate quotient

1580

1580

participants received Drug A and did not recover from their condition.

Closure

Detecting if an Email Is Spam

Like Venn diagrams and frequency tables, tree diagrams relate the probability of a conditional event to a subset of the event occurring. With this in mind, reconsider the example given at the beginning of the lesson. These three points about spam emails are known.

$50 %$ of emails are spam.
$15 %$ of spam emails contain the word free.
$1 %$ of non-spam emails contain the word free.

Jordan gets an email with the word free in it.

a Draw a tree diagram to visualize the situation.

b What is the probability that the email is spam? If necessary, round the answer to the two decimal places.

Answer

a Example Tree Diagram:

P (S ∣ C) \approx 0.94

Hint

a Start by finding the probabilities that will be written on branches.

b Determine the paths of the tree diagram that lead to emails that contains the word free.

Solution

a Start by defining the events.

Therefore,

P (S),

P (C ∣ S),

and

P (C ∣ N S)

are given.

P (S) P (C ∣ S) P (C ∣ N S) = 50 %, or 0.5 = 15 %, or 0.15 = 11 %, or 0.01

Notice that

N S

is the complement of

S

and

N C

is the complement of

C .

Recall that the probability of the complement of an event is

1

minus the probability of the event. Therefore, the probabilities of the complements of the given events can be found.

P (S) 0.5 P (C ∣ S) 0.15 P (C ∣ N S) 0.01 \Leftrightarrow \Leftrightarrow \Leftrightarrow P (N S) 1 - 0.5 = 0.5 P (N C ∣ S) 1 - 0.15 = 0.85 P (N C ∣ N S) 1 - 0.01 = 0.99

With the events and probabilities determined, a tree diagram can be drawn as shown.

b Jordan gets an email that contains the word free. The probability that this email is spam needs to be calculated.

P (S ∣ C) = ?

To find it, the tree diagram drawn in the previous part will be used. Consider all the paths that lead to event

C,

the event that an email contains the word free.

The probability of either of these two paths occurring is the sum of the products of the probabilities on the paths.

0.5 \cdot 0.15 + 0.5 \cdot 0.01 = 0.08

Of these two paths, the topmost one leads to the favorable event. The probability of this event is the product of the probabilities on the path.

0.5 \cdot 0.15 = 0.075

As a result, the ratio of the probability of the favorable path to the probability of either of the possible paths gives

P (S ∣ C) .

P (S ∣ C) = \frac{0 . 0 7 5}{0 . 0 8}

CalcQuot

Calculate quotient

P (S ∣ C) = 0.9375

RoundDec

Round to $2$ decimal place(s)

P (S ∣ C) \approx 0.94

An email that contains the word free has a

0.94

probability that it is a spam email.

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Catch-Up and Review

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Solution

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