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:

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

2.0What is Conditional Probability?

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

Notation:

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

Formula:

$P(A\mid B)=\frac{P(A\cap B)}{P(B)},\text{ if }P(B)\ne 0$

Also Practice: Conditional Probability Questions

3.0What is the Difference Between Probability and Conditional Probability?

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

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

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

$P(\text{ Red }\mid \text{ Not blue })=\frac{3}{7}$

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:

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

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

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:

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

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:

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

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 → $P\left(A_1\right)=\frac{1}{2}$
• $A_2$ : Choosing Urn II → $P\left(A_2\right)=\frac{1}{2}$

Now,

• $P\left(R\mid A_1\right)=\frac{3}{8},$
• $P\left(R\mid A_2\right)=\frac{4}{6}=\frac{2}{3}$

By Bayes' Theorem:

$P\left(A_2\mid R\right)=\frac{P\left(A_2\right)\cdot P\left(R\mid A_2\right)}{P\left(A_1\right)\cdot P\left(R\mid A_1\right)+P\left(A_2\right)\cdot P\left(R\mid A_2\right)}$

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

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

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 $P(A\mid B)$ and state whether A and B are independent.

Solution:

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

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

Answer: $P(A\mid B)=0.4$, 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:

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

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,

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

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

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