Is conditional probability always less than probability?

Not necessarily. It depends on how the given event (condition) affects the likelihood of the desired event.

Can two events be independent in conditional probability?

Yes. If P(A|B)=P(A) , then A and B are independent events.

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Probability and Conditional Probability Examples

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Probability and Conditional Probability

Q: What is the difference between probability and conditional probability?

Probability is the general chance of an event. Conditional probability is the chance of an event occurring under a specific condition or given another event has happened.

Probability is a key concept in mathematics that helps us measure the likelihood of an event occurring. However, when one event is known to have occurred and we want to calculate the probability of another event, we use conditional probability. Let’s explore how to define probability and conditional probability, understand their differences, and solve problems involving both.

1.0What is Probability?

Probability is the measure of the likelihood that a particular event will occur. It ranges from 0 to 1, where:

0 means the event is impossible
1 means the event is certain

Formula:

$P (E) = \frac{Number of favorable outcomes}{Total number of outcomes}$

2.0What is Conditional Probability?

Conditional probability is the probability of one event occurring given that another event has already occurred.

Notation:

$P (A ∣ B)$ (read as "Probability of A given B")

Formula:

$P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}, if P (B) \neq = 0$

Also Practice: Conditional Probability Questions

3.0What is the Difference Between Probability and Conditional Probability?

Basis	Probability	Conditional Probability
Definition	Likelihood of an event	Likelihood of an event given another event has occurred
Symbol	P(A)	P(A\|B)
Dependency	Independent	Depends on a known event
Example	Rolling a 3 on a die	Rolling a 3 given it is an odd number

4.0Why is Conditional Probability Important?

It is crucial when:

Events are dependent
We need to update probabilities after observing some information
Applications in Bayes’ Theorem, machine learning, risk analysis, etc.

5.0Probability and Conditional Probability Solved Problems

Example 1: A card is drawn from a standard deck of 52 cards. What is the probability that it is a king given that it is a face card?

Solution:

Total face cards = Jack, Queen, King × 4 = 12
Favorable = 4 Kings

$P (King ∣ Face card) = \frac{4}{12} = \frac{1}{3}$

Example 2: A bag contains 3 red, 4 green, and 2 blue balls. A ball is drawn at random.
What is the probability that the ball is red given that it is not blue?

Solution:

Total balls = 3 + 4 + 2 = 9
Non-blue = 3 + 4 = 7
Red balls = 3

$P (Red ∣ Not blue) = \frac{3}{7}$

Example 3: Two cards are drawn one after another from a deck of 52 cards without replacement. What is the probability that both are aces?

Solution:

$P (1 st Ace) = \frac{4}{52}$
$P (2 nd Ace ∣ 1 st Ace) = \frac{3}{51}$

$P (Both Aces) = \frac{4}{52} \times \frac{3}{51} = \frac{1}{221}$

Example 4: In a class of 60 students, 36 like Maths, 30 like Science, and 20 like both.

(a) What is the probability a student likes Maths?

(b) What is the probability a student likes Maths given they like Science?

Solution:

Let:

$P (M) = \frac{36}{60} = 0.6$
$P (S) = \frac{30}{60} = 0.5$
$P (M \cap S) = \frac{20}{60} = \frac{1}{3}$

(b) Conditional Probability:

$P (M ∣ S) = \frac{P ( M \cap S )}{P ( S )} = \frac{1/3}{0.5} = \frac{2}{3}$

Example 5: Two cards are drawn without replacement from a deck of 52 cards. Find the probability that the second card is a heart, given that the first card was a spade.

Solution:

Since cards are drawn without replacement:

After removing a spade, 51 cards remain.
Hearts are unaffected: 13 hearts still remain.

So:

$P (2 nd is Heart ∣ 1 st is Spade) = \frac{13}{51}$

Answer: $\frac{13}{51}$

Example 6: Urn I contains 3 red and 5 black balls. Urn II contains 4 red and 2 black balls. An urn is chosen at random, and a ball is drawn. If the ball is red, what is the probability it was drawn from Urn II?

Solution:

Let:

$A_{1}$ : Choosing Urn I → $P (A_{1}) = \frac{1}{2}$
$A_{2}$ : Choosing Urn II → $P (A_{2}) = \frac{1}{2}$

Now,

$P (R ∣ A_{1}) = \frac{3}{8},$
$P (R ∣ A_{2}) = \frac{4}{6} = \frac{2}{3}$

By Bayes' Theorem:

$P (A_{2} ∣ R) = \frac{P ( A _{2} ) \cdot P ( R ∣ A _{2} )}{P ( A _{1} ) \cdot P ( R ∣ A _{1} ) + P ( A _{2} ) \cdot P ( R ∣ A _{2} )}$

$= 12 \cdot 2312 \cdot 38 + 12 \cdot 23 = 12 \cdot 23316 + 13 = 12 \cdot 232548 = 132548 = 4875 = 1625$

$= \frac{\frac{1}{2} \cdot \frac{2}{3}}{\frac{1}{2} \cdot \frac{3}{8} + \frac{1}{2} \cdot \frac{2}{3}} = \frac{\frac{1}{2} \cdot \frac{2}{3}}{\frac{3}{16} + \frac{1}{3}} = \frac{\frac{1}{2} \cdot \frac{2}{3}}{\frac{25}{48}} = \frac{\frac{1}{3}}{\frac{25}{48}} = \frac{48}{75} = \frac{16}{25}$

Answer: $\frac{16}{25}$

Example 7: Events A and B are such that P(A) = 0.4, P(B) = 0.5, and $P (A \cap B) = 0.2$ . Find $P (A ∣ B)$ and state whether A and B are independent.

Solution:

$P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )} = \frac{0.2}{0.5} = 0.4$

Since $P (A ∣ B) = P (A)$ , events A and B are independent.

Answer: $P (A ∣ B) = 0.4$ , A and B are independent

Example 8: A die is thrown until a 6 appears. What is the probability that a 6 appears on the third throw?

Solution:

This is a geometric probability problem:

Success = getting a 6 → $p = \frac{1}{6}$
Failure = not getting a 6 → $q = \frac{5}{6}$

To get a 6 on the third throw, first two must be failures, third a success:

$P (6 on 3rd throw) = (\frac{5}{6})^{2} \cdot \frac{1}{6} = \frac{25}{216}$

Answer: $\frac{25}{216}$

Example 9: A number is selected at random from 1 to 100. Find the probability that it is divisible by 5, given that it is divisible by 2.

Solution:

Let:

Total numbers divisible by 2 = 50 (even numbers from 2 to 100)
Numbers divisible by both 2 and 5 = Numbers divisible by 10 = 10, 20, ..., 100 → 10 numbers

So,

$P (Divisible by 5 ∣ Divisible by 2) = \frac{10}{50} = \frac{1}{5}$

Answer: $\frac{1}{5}$

Also Read:

_{Probability and Statistics}	_{Empirical Probability}	_{Total Probability Theorem}
_{Multiplication Theorem on Probability}	_{Probability and Statistics questions with solutions}	_Statistics

Join ALLEN!

(Session 2026 - 27)

Name

Mobile Number

Class

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Preferred Mode

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I agree to

I authorise ALLEN Career Institute Pvt Ltd to send me regular updates via Phone calls, Whatsapp, SMS, Robocalls (Automated Calls), Emails, or on postal address.

Probability and Conditional Probability

Probability is a key concept in mathematics that helps us measure the likelihood of an event occurring. However, when one event is known to have occurred and we want to calculate the probability of another event, we use conditional probability. Let’s explore how to define probability and conditional probability, understand their differences, and solve problems involving both.

1.0What is Probability?

Probability is the measure of the likelihood that a particular event will occur. It ranges from 0 to 1, where:

0 means the event is impossible
1 means the event is certain

Formula:

$P (E) = \frac{Number of favorable outcomes}{Total number of outcomes}$

2.0What is Conditional Probability?

Conditional probability is the probability of one event occurring given that another event has already occurred.

Notation:

$P (A ∣ B)$ (read as "Probability of A given B")

Formula:

$P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}, if P (B) \neq = 0$

Also Practice: Conditional Probability Questions

3.0What is the Difference Between Probability and Conditional Probability?

Basis	Probability	Conditional Probability
Definition	Likelihood of an event	Likelihood of an event given another event has occurred
Symbol	P(A)	P(A\|B)
Dependency	Independent	Depends on a known event
Example	Rolling a 3 on a die	Rolling a 3 given it is an odd number

4.0Why is Conditional Probability Important?

It is crucial when:

Events are dependent
We need to update probabilities after observing some information
Applications in Bayes’ Theorem, machine learning, risk analysis, etc.

5.0Probability and Conditional Probability Solved Problems

Example 1: A card is drawn from a standard deck of 52 cards. What is the probability that it is a king given that it is a face card?

Solution:

Total face cards = Jack, Queen, King × 4 = 12
Favorable = 4 Kings

$P (King ∣ Face card) = \frac{4}{12} = \frac{1}{3}$

Example 2: A bag contains 3 red, 4 green, and 2 blue balls. A ball is drawn at random.
What is the probability that the ball is red given that it is not blue?

Solution:

Total balls = 3 + 4 + 2 = 9
Non-blue = 3 + 4 = 7
Red balls = 3

$P (Red ∣ Not blue) = \frac{3}{7}$

Example 3: Two cards are drawn one after another from a deck of 52 cards without replacement. What is the probability that both are aces?

Solution:

$P (1 st Ace) = \frac{4}{52}$
$P (2 nd Ace ∣ 1 st Ace) = \frac{3}{51}$

$P (Both Aces) = \frac{4}{52} \times \frac{3}{51} = \frac{1}{221}$

Example 4: In a class of 60 students, 36 like Maths, 30 like Science, and 20 like both.

(a) What is the probability a student likes Maths?

(b) What is the probability a student likes Maths given they like Science?

Solution:

Let:

$P (M) = \frac{36}{60} = 0.6$
$P (S) = \frac{30}{60} = 0.5$
$P (M \cap S) = \frac{20}{60} = \frac{1}{3}$

(b) Conditional Probability:

$P (M ∣ S) = \frac{P ( M \cap S )}{P ( S )} = \frac{1/3}{0.5} = \frac{2}{3}$

Example 5: Two cards are drawn without replacement from a deck of 52 cards. Find the probability that the second card is a heart, given that the first card was a spade.

Solution:

Since cards are drawn without replacement:

After removing a spade, 51 cards remain.
Hearts are unaffected: 13 hearts still remain.

So:

$P (2 nd is Heart ∣ 1 st is Spade) = \frac{13}{51}$

Answer: $\frac{13}{51}$

Example 6: Urn I contains 3 red and 5 black balls. Urn II contains 4 red and 2 black balls. An urn is chosen at random, and a ball is drawn. If the ball is red, what is the probability it was drawn from Urn II?

Solution:

Let:

$A_{1}$ : Choosing Urn I → $P (A_{1}) = \frac{1}{2}$
$A_{2}$ : Choosing Urn II → $P (A_{2}) = \frac{1}{2}$

Now,

$P (R ∣ A_{1}) = \frac{3}{8},$
$P (R ∣ A_{2}) = \frac{4}{6} = \frac{2}{3}$

By Bayes' Theorem:

$P (A_{2} ∣ R) = \frac{P ( A _{2} ) \cdot P ( R ∣ A _{2} )}{P ( A _{1} ) \cdot P ( R ∣ A _{1} ) + P ( A _{2} ) \cdot P ( R ∣ A _{2} )}$

$= 12 \cdot 2312 \cdot 38 + 12 \cdot 23 = 12 \cdot 23316 + 13 = 12 \cdot 232548 = 132548 = 4875 = 1625$

$= \frac{\frac{1}{2} \cdot \frac{2}{3}}{\frac{1}{2} \cdot \frac{3}{8} + \frac{1}{2} \cdot \frac{2}{3}} = \frac{\frac{1}{2} \cdot \frac{2}{3}}{\frac{3}{16} + \frac{1}{3}} = \frac{\frac{1}{2} \cdot \frac{2}{3}}{\frac{25}{48}} = \frac{\frac{1}{3}}{\frac{25}{48}} = \frac{48}{75} = \frac{16}{25}$

Answer: $\frac{16}{25}$

Example 7: Events A and B are such that P(A) = 0.4, P(B) = 0.5, and $P (A \cap B) = 0.2$ . Find $P (A ∣ B)$ and state whether A and B are independent.

Solution:

$P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )} = \frac{0.2}{0.5} = 0.4$

Since $P (A ∣ B) = P (A)$ , events A and B are independent.

Answer: $P (A ∣ B) = 0.4$ , A and B are independent

Example 8: A die is thrown until a 6 appears. What is the probability that a 6 appears on the third throw?

Solution:

This is a geometric probability problem:

Success = getting a 6 → $p = \frac{1}{6}$
Failure = not getting a 6 → $q = \frac{5}{6}$

To get a 6 on the third throw, first two must be failures, third a success:

$P (6 on 3rd throw) = (\frac{5}{6})^{2} \cdot \frac{1}{6} = \frac{25}{216}$

Answer: $\frac{25}{216}$

Example 9: A number is selected at random from 1 to 100. Find the probability that it is divisible by 5, given that it is divisible by 2.

Solution:

Let:

Total numbers divisible by 2 = 50 (even numbers from 2 to 100)
Numbers divisible by both 2 and 5 = Numbers divisible by 10 = 10, 20, ..., 100 → 10 numbers

So,

$P (Divisible by 5 ∣ Divisible by 2) = \frac{10}{50} = \frac{1}{5}$

Answer: $\frac{1}{5}$

Also Read:

_{Probability and Statistics}	_{Empirical Probability}	_{Total Probability Theorem}
_{Multiplication Theorem on Probability}	_{Probability and Statistics questions with solutions}	_Statistics

Frequently Asked Questions

What is the difference between probability and conditional probability?

Is conditional probability always less than probability?

Can two events be independent in conditional probability?

Frequently Asked Questions

What is the difference between probability and conditional probability?

Is conditional probability always less than probability?

Can two events be independent in conditional probability?

Join ALLEN!

Probability and Conditional Probability

1.0What is Probability?

2.0What is Conditional Probability?

Notation:

Formula:

3.0What is the Difference Between Probability and Conditional Probability?

4.0Why is Conditional Probability Important?

5.0Probability and Conditional Probability Solved Problems

Table of Contents

Join ALLEN!

Probability and Conditional Probability

1.0What is Probability?

2.0What is Conditional Probability?

Notation:

Formula:

3.0What is the Difference Between Probability and Conditional Probability?

4.0Why is Conditional Probability Important?

5.0Probability and Conditional Probability Solved Problems

Table of Contents