What Is the Difference Between Conditional and Joint Probability?

Conditional probability is the probability of one event given that another has occurred. Joint probability is the probability of both events occurring together.

Can Conditional Probability Be Used for Independent Events?

Yes. For independent events, P(A|B) = P(A), as the occurrence of B doesn’t affect A.

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Conditional Probability Questions

Conditional probability is a concept in probability theory that calculates the likelihood of an event occurring, given that another event has already occurred. It’s denoted as P(A|B), representing the probability of event A occurring, assuming event B is true. Conditional probability helps in understanding how one event influences the probability of another, especially in dependent events, and is fundamental in statistics and data analysis.

1.0What is Conditional Probability?

Before diving into the questions, let's quickly define conditional probability. The conditional probability of an event A, given that event B has occurred, is expressed as:

Formula

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

Where:

P(A|B) indicates the probability of event A occurring, assuming event B has already happened.
P(A ∩ B) refers to the probability that both events A and B occur.
P(B) refers to the probability that event B occurs.

2.0Solved Examples On Conditional Probability

Example 1: In a deck of 52 cards, what is the probability of drawing a King, given that the card drawn is a face card (Jack, Queen, or King)?

Solution:

The total number of face cards in a deck is 12 (4 Jacks, 4 Queens, and 4 Kings).
The number of Kings is 4.

We want to find the conditional probability P(King | Face Card).

Apply Conditional Probability formula :-

$P (K in g ∣ F a ce C a r d) = \frac{P ( K in g \cap F a ce C a r d )}{P ( F a ce C a r d )}$ $= \frac{\frac{4}{52}}{\frac{12}{52}} = \frac{4}{12} = \frac{1}{3}$

Thus, the probability of drawing a King, given that the card is a face card, is 1/3.

Example 2: Roll a fair die twice. Let A be the event that the sum of the two rolls equals six, and let B be the event that the same number comes up twice. What is P(A/B)?

(A) 1/6 (B) 5/36 (C) 1/5 (D) none

Solution:

A={(1, 5), (4, 4), (3, 3), (4, 2), (5, 1)}

B={(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}

Apply Conditional Probability formula :-

$P (\frac{A}{B}) = \frac{P ( A \cap B )}{P ( B )} = \frac{n ( A \cap B )}{n ( B )} = \frac{1}{6}$

Example 3: In a class, 30% of the students failed in Physics, 25% failed in Mathematics and 15% failed in both Physics and Mathematics. If a student is selected at random failed in Mathematics, find the probability that he failed in Physics also.

Solution:

Let A be the event "failed in Physics" and B be the event "failed in Mathematics". We want to find $P (\frac{A}{B})$ .

It is given that P(A) = $\frac{300}{100}$ and P(B) = $\frac{25}{100}$

Also P(A∩B)= $\frac{15}{100}$

Therefore $P (\frac{A}{B}) = \frac{P ( A \cap B )}{P ( B )}$ $\frac{( \frac{15}{100} )}{( \frac{25}{100} )}$ $= \frac{15}{25} = \frac{3}{5}$

Using conditional probability formula

Example 4: If a pair of dice is thrown and it is known that sum of the numbers is even, then find the probability that the sum is less than 6.

Solution:

Let A be the given event and let B be the event, whose probability is to be found. Then

Required probability.

$P (\frac{B}{A}) = \frac{P ( B \cap A )}{P ( A )} = \frac{\frac{4}{36}}{\frac{18}{36}} = \frac{2}{9}$

Example 5: If 2 numbers appearing on the dice are different, then find probability that

(i) Their sum is six.

(ii) One face is 1.

(iii) Sum exceeds 9.

(iv) Sum is even

Solution:

(1, 5), (2, 4), (4, 2), (5, 1)

n(F)=4, n(s)=30

$P = \frac{4}{30}$

(1,2),(2,1),(1,3),(3,1), (1,4), (4,1), (1,5), (5,1), (1,6), (6,1)

n(F)=10, n(s)=30

$P = \frac{1}{3}$

(4,6), (6,4), (6,5), (5,6)

n(F)=4, n(s)=30

$P = \frac{4}{30}$

(1,3), (2,4), (3,1), (4,2), (5,1), (1,5), (2,6), (3,5), (4,6), (5,3), (6,2), (6,4)

n(F)=12

P= $\frac{12}{30} = \frac{2}{5}$

3.0Sample Questions on Conditional Probability

What is Conditional Probability?

Ans: Conditional probability refers to the likelihood of an event happening, given that another event has already taken place. It is represented as. $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$

How Do You Calculate Conditional Probability?

Ans: Use the formula $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ , where $P (A \cap B)$ is the probability of both events occurring, and P(B) is the probability of event B.

What Is Bayes’ Theorem?

Ans: Bayes’ Theorem helps calculate the probability of an event based on prior knowledge of related events. It's given by . $P (A ∣ B) = \frac{P ( B ∣ A ) . P ( A )}{P ( B )}$

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Conditional Probability Questions

Conditional probability is a concept in probability theory that calculates the likelihood of an event occurring, given that another event has already occurred. It’s denoted as P(A|B), representing the probability of event A occurring, assuming event B is true. Conditional probability helps in understanding how one event influences the probability of another, especially in dependent events, and is fundamental in statistics and data analysis.

1.0What is Conditional Probability?

Before diving into the questions, let's quickly define conditional probability. The conditional probability of an event A, given that event B has occurred, is expressed as:

Formula

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

Where:

P(A|B) indicates the probability of event A occurring, assuming event B has already happened.
P(A ∩ B) refers to the probability that both events A and B occur.
P(B) refers to the probability that event B occurs.

2.0Solved Examples On Conditional Probability

Example 1: In a deck of 52 cards, what is the probability of drawing a King, given that the card drawn is a face card (Jack, Queen, or King)?

Solution:

The total number of face cards in a deck is 12 (4 Jacks, 4 Queens, and 4 Kings).
The number of Kings is 4.

We want to find the conditional probability P(King | Face Card).

Apply Conditional Probability formula :-

$P (K in g ∣ F a ce C a r d) = \frac{P ( K in g \cap F a ce C a r d )}{P ( F a ce C a r d )}$ $= \frac{\frac{4}{52}}{\frac{12}{52}} = \frac{4}{12} = \frac{1}{3}$

Thus, the probability of drawing a King, given that the card is a face card, is 1/3.

Example 2: Roll a fair die twice. Let A be the event that the sum of the two rolls equals six, and let B be the event that the same number comes up twice. What is P(A/B)?

(A) 1/6 (B) 5/36 (C) 1/5 (D) none

Solution:

A={(1, 5), (4, 4), (3, 3), (4, 2), (5, 1)}

B={(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}

Apply Conditional Probability formula :-

$P (\frac{A}{B}) = \frac{P ( A \cap B )}{P ( B )} = \frac{n ( A \cap B )}{n ( B )} = \frac{1}{6}$

Example 3: In a class, 30% of the students failed in Physics, 25% failed in Mathematics and 15% failed in both Physics and Mathematics. If a student is selected at random failed in Mathematics, find the probability that he failed in Physics also.

Solution:

Let A be the event "failed in Physics" and B be the event "failed in Mathematics". We want to find $P (\frac{A}{B})$ .

It is given that P(A) = $\frac{300}{100}$ and P(B) = $\frac{25}{100}$

Also P(A∩B)= $\frac{15}{100}$

Therefore $P (\frac{A}{B}) = \frac{P ( A \cap B )}{P ( B )}$ $\frac{( \frac{15}{100} )}{( \frac{25}{100} )}$ $= \frac{15}{25} = \frac{3}{5}$

Using conditional probability formula

Example 4: If a pair of dice is thrown and it is known that sum of the numbers is even, then find the probability that the sum is less than 6.

Solution:

Let A be the given event and let B be the event, whose probability is to be found. Then

Required probability.

$P (\frac{B}{A}) = \frac{P ( B \cap A )}{P ( A )} = \frac{\frac{4}{36}}{\frac{18}{36}} = \frac{2}{9}$

Example 5: If 2 numbers appearing on the dice are different, then find probability that

(i) Their sum is six.

(ii) One face is 1.

(iii) Sum exceeds 9.

(iv) Sum is even

Solution:

(1, 5), (2, 4), (4, 2), (5, 1)

n(F)=4, n(s)=30

$P = \frac{4}{30}$

(1,2),(2,1),(1,3),(3,1), (1,4), (4,1), (1,5), (5,1), (1,6), (6,1)

n(F)=10, n(s)=30

$P = \frac{1}{3}$

(4,6), (6,4), (6,5), (5,6)

n(F)=4, n(s)=30

$P = \frac{4}{30}$

(1,3), (2,4), (3,1), (4,2), (5,1), (1,5), (2,6), (3,5), (4,6), (5,3), (6,2), (6,4)

n(F)=12

P= $\frac{12}{30} = \frac{2}{5}$

3.0Sample Questions on Conditional Probability

What is Conditional Probability?

Ans: Conditional probability refers to the likelihood of an event happening, given that another event has already taken place. It is represented as. $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$

How Do You Calculate Conditional Probability?

Ans: Use the formula $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ , where $P (A \cap B)$ is the probability of both events occurring, and P(B) is the probability of event B.

What Is Bayes’ Theorem?

Ans: Bayes’ Theorem helps calculate the probability of an event based on prior knowledge of related events. It's given by . $P (A ∣ B) = \frac{P ( B ∣ A ) . P ( A )}{P ( B )}$

Frequently Asked Questions

What Is the Difference Between Conditional and Joint Probability?

Can Conditional Probability Be Used for Independent Events?

Join ALLEN!

Conditional Probability Questions

1.0What is Conditional Probability?

2.0Solved Examples On Conditional Probability

3.0Sample Questions on Conditional Probability

Table of Contents

Frequently Asked Questions

What Is the Difference Between Conditional and Joint Probability?

Can Conditional Probability Be Used for Independent Events?

Join ALLEN!

Conditional Probability Questions

1.0What is Conditional Probability?

2.0Solved Examples On Conditional Probability

3.0Sample Questions on Conditional Probability

Table of Contents