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?

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.

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Catch-Up and Review

Hint

Solution

Answer

Hint

Solution