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

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

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

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

A :

Passenger Survived

B :

First Class Passenger

C :

Second Class Passenger

D :

Third Class Passenger

	$A :$ 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, $P (A ∣ D) \neq = P (A)$

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)

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

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

	$A :$ 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, $P (A ∣ D) \neq = P (A)$

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)

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

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

The following table shows two events. Event $A$ is colored in red and Event $B$ in blue.

sample space showing the result of rolling two dices

Determine

P (B ∣ A) .

Write the answer as a fraction in its simplest form.

We want to determine the probability of B happening given that A has happened. Before that, we need to define the events.

Defining the Events

As we can see, event A covers all the possible outcomes that contain at least one die showing a 3. Event B covers all outcomes where the sum of the dice is 9. We get the following events. A: &The rolled dice show at least one3. B: &The dice add up to9. To determine P(B|A), we can use the following formula. P(B|A)=P(A and B)/P(A) We can determine the probabilities on the right-hand side using the given diagram.

Determining the Probabilities

Examining the diagram, we can count the number of outcomes for each event. Be careful not to double count the 6.

Now, we can define the probabilities we need. As we can see, 11 out of the 36 total outcomes result in Event A happening. As for Event B, there are four favorable outcomes. However, for both A and B to happen, we only have two favorable outcomes. We get the following probabilities. P(A)& =11/36 [0.7em] P(A and B)& =2/36

Calculating P(B|A)

Finally, we can calculate the probability we are asked.

The probability of P(B|A) is 211.

Consider the following Venn diagram for the two events $A$ and $B .$

The probability of

B

given

A

is equal to

0.2 .

P (B ∣ A) = 0.2

How many observations belong to only

B ?

The conditional probability P(B|A) can be written as the following formula. P(B|A)=P(A and B)/P(A) This formula divides the probability of A and B by the probability of A. However, we can also find P(B|A) by dividing the number of observations in A and B by the total number of observations in A. P(B|A)=n(A and B)/n(A) Let's visualize the situation.

Since we know that P(B|A)=0.2, we can solve for n(A) in our equation.

As we can see, the number of observations in A, n(A), is 80. This means the observations in A but not B must be 80-16=64. Let's add this to the Venn diagram.

Since 64 corresponds to 50 % of the observations, we know that the total number of observations in the entire sample space, n(S), must be 64* 2=128. Now we can find the number of observations that belong to B but not A by calculating 25 % of the total number of observations.

There are 32 observations that belong to event B but not event A.

The following diagrams show a study on whether students at a school play an instrument or play sports. However, both diagrams are incomplete.

representation of the survey with a table and a Venn diagram

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

P (A ∣ B)

P (B ∣ A)

Notice that both the table and the Venn diagram have insufficient information on their own. However, when they are both considered, we can work out the missing information. According to the Venn diagram, 16 students play both an instrument and a sport.

This corresponds to the top left cell of the frequency table, (Yes,Yes).

From the Venn diagram, we also know that 80 students took the survey. This number should equal the sum of the four cells in the table. Since we know the count of three of the cells, we can determine the count in the fourth cell by subtracting them from 80. 80-16-30-9=25 We can now complete both the Venn diagram and the table.

To calculate P(A|B), we need to find P(AandB) and P(B). These values can be found from the diagram. P(AandB) = 16/80 [0.7em] P(B) = 46/80 Finally, we can calculate P(A|B).

In the previous section we identified the missing information in the diagrams. Let's remind ourselves of that.

To calculate P(B|A), we need to find P(AandB) and P(A). These values can be found from the diagram. P(AandB) = 16/80 [0.7em] P(A) = 41/80 Let's substitute these values into the formula for the conditional probability.

Conditional Probability in the Uniform Probability Model

Catch-Up and Review

Detecting if an Email Is Spam

Reviewing the Intuition Behind the Conditional Probability Formula

Using a Venn Diagram to Find Conditional Probability

Hint

Solution

Practice Finding Conditional Probability

Using a Table to Find Conditional Probabilities

Answer

Hint

Solution

Practice Finding Conditional Probability

Using a Tree Diagram to Find Conditional Probabilities

Answer

Hint

Solution

Detecting if an Email Is Spam

Answer

Hint

Solution

Defining the Events

Determining the Probabilities

Calculating P(B|A)

Conditional Probability in the Uniform Probability Model

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