Question

A die is thrown three times. Events A and B are defined as below:
A: 6 on the third throw
B: 4 on the first and 5 on the second throw
The probability of A given that B has already occurred, is:

• A. $\frac{1}{6}$
• B. $\frac{2}{3}$
• C. $\frac{3}{4}$
• D. $\frac{1}{2}$

Answer 1

The Correct Option is A

To address this, we will compute the conditional probability of event A given event B, denoted as \( P(A|B) \). The formula for conditional probability is:

\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]

Step 1: Compute \( P(B) \)

Event B occurs with a 4 on the first throw and a 5 on the second. The probability of rolling any specific number on a fair six-sided die is \(\frac{1}{6}\). Therefore,

\[ P(B) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} \]

Step 2: Compute \( P(A \cap B) \)

Event \( A \cap B \) occurs when event B is supplemented by a 6 on the third throw. This is calculated as:

\[ P(A \cap B) = P(4 \text{ on first}) \times P(5 \text{ on second}) \times P(6 \text{ on third}) \]

\[ = \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{216} \]

Step 3: Compute \( P(A|B) \)

Substitute the derived probabilities into the conditional probability formula:

\[ P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{216}}{\frac{1}{36}} \]

Simplification by multiplying by the reciprocal yields:

\[ = \frac{1}{216} \times \frac{36}{1} = \frac{1}{6} \]

The conditional probability of event A given B is \(\frac{1}{6}\). The correct answer is therefore \(\frac{1}{6}\).