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 solve this problem, we'll calculate the conditional probability of event A given that event B has occurred. We need to determine \( P(A|B) \). The formula for conditional probability is:

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

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

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

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

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

Event \( A \cap B \) occurs when a 6 is rolled on the third throw in addition to event B. For this,

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

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

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

Substitute the values found into the conditional probability formula:

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

To simplify, multiply by the reciprocal:

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

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

Answer 2

Event \(B\) indicates that the first throw is a 4 and the second throw is a 5. The probability of event \(B\) occurring is independent of event \(A\). Since the die is fair, the probability of rolling a 6 on any throw is \(\frac{1}{6}\).

Therefore, the probability of \(A\) given that \(B\) has already occurred is:

\[ P(A|B) = P(A) = \frac{1}{6}. \]