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 Venn diagram shows the percentages of households in a village that subscribe to the morning and evening papers.

If a household is selected at random, what is the probability that the household subscribes to at least one of the papers?

Vincenzo subscribes to one paper. What is the probability that he is subscribed to the morning paper? Round to the nearest percent.

What is the probability that Vincenzo's neighbor Ali also subscribes to the evening paper if it is known that he is subscribed to the morning paper? Round to the nearest percent.

We have been given the percentages that subscribe to one or both of the papers. Only Morning Paper: & 40 % Only Evening Paper: & 13 % Morning and Evening Papers: & 25 % If we add these percentages, we obtain how many percent of the village that subscribe to at least one of the papers. 40 % +25 % +13 % =78 % As we can see, 78 % of all households subscribe to at least one of the papers.

We know that Vincenzo subscribes to a paper. In other words, it is given that he is in the 78 % that we calculated in in the previous exercise. We want to know the probability that he is subscribed to only the morning paper. Therefore, we want to calculate the following conditional probability. P(morning paper|subscriber) We can calculate this by dividing the probability that a household that is a subscriber to the paper subscribes to morning paper by the probability that a household is a subscriber. P(morning paper and subscriber)/P(subscriber) We can illustrate this on the Venn diagram below.

If we divide the favorable outcomes by the size of the population, we can determine the conditional probability we are asked to find.

The probability that Vincenzo subscribes only to the morning paper, given that he is a subscriber, is about 51 %.

This time we want to know the conditional probability of Ali subscribing to the evening paper given that he also subscribes to the morning paper. P(morning paper and evening paper)/P(morning paper) Let's illustrate the population and favorable outcomes in the diagram.

Now we can calculate the conditional probability.

The probability that Ali is subscribed to the evening paper, given that he is also a subscriber to the morning paper, is about 38 %.

Event $A$ and $B$ are dependent. Calculate $P (A)$ given the following probabilities. Answer with a decimal number.

P (A and B) = 0.1 P (B ∣ A) = 0.4

P (A and B) = 0.3 P (B ∣ A) = 0.5

To solve for P(A), we will use the following formula. P(B|A)=P(A and B)/P(A) We have been given P(B|A) and P(A and B). If we substitute these values into the formula, we can solve for P(A).

The probability of A is 0.25.

As in the previous part, we will substitute the given probabilities into the formula and solve for P(A).

The probability of A is 0.6.

A bag holds three rolled-up Donald Duck comics and seven rolled-up Archie comics. Zain picks a random comic for themselves and one for their little brother to read. The first one is a Donald Duck and the second one is an Archie comic. Their older sister Zosia, who is in high school, attempts to calculate the probability of this event happening.

Calculations on conditional probability that are wrong

She has made a mistake in her calculations. Describe her mistake and calculate the correct probability. Round the answer to the nearest percent.

From the exercise, we know that Zain first picks a Donald Duck comic for themselves and then an Archie comic for their little brother. Since there is one less comic in the bag after the first one is picked, there is one less comic available in the bag. r|r & Number [0.5em] [-0.5em] Donald Duck Comics: & 3 - 1 2 [0.5em] Archie Comics: & 7 7 [0.5em] Total:& 10 - 1 9 Therefore, the probability of picking an Archie will be 7 9.

Now, by using the formula for the conditional probability, we can find P(Archie&Donald Duck).

The probability of choosing a Donald Duck comic first and then an Archie comic is about 23 %. With this information, the calculations can be corrected.

At a college,

80 %

of the students eat lunch at the local cafeteria. Given that a student eats lunch at the cafeteria, there is a

1

5

probability that the student sometimes also makes their own food. What percentage of students who eat at the cafeteria also occasionally make their own food? Answer as a percentage.

Let's label the events of eating at the cafeteria and making own food. C: & Student eats at the cafeteria. F: & Student makes their own food. We want to know the probability that a student who eats at the cafeteria also occasionally makes their own food, P(C andF). We can write it using the formula for the conditional probability. P(F|C) = P(CandF)/P(C) ⇕ P(CandF) = P(C) * P(F|C) From the exercise, we know that 80 % of all students eat at the cafeteria. Furthermore, we are given the probability of a student making their own food, given that they eat at the cafeteria, as 15. P(C) & = 0.8 P(F|C)& = 1/5 With this information, we can calculate P(CandF) by substituting these values into the equation.

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

Conditional Probability in the Uniform Probability Model

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	9 Theory slides
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	Each lesson is meant to take 1-2 classroom sessions