Question

A die is rolled twice. Let $A$ be an event of getting 1 in the first roll and let $B$ be an event of getting 4 in the second roll. Then $P(A|B)$ is equal to

• A. $\frac{1}{6}$
• B. $\frac{1}{3}$
• C. $\frac{1}{9}$
• D. $\frac{1}{18}$
• E. $\frac{1}{36}$

Hint:

For independent successive trials (like dice rolls or coin tosses), the conditional probability of an event given another trial's outcome is just the probability of the event itself.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We need to find the conditional probability of event A given that event B has occurred, denoted as \(P(A|B)\).
Since rolling a die twice constitutes independent events, the outcome of the second roll does not affect the first roll.
Step 2: Key Formula or Approach:
For independent events, \(P(A|B) = P(A)\).
Alternatively, use the definition of conditional probability: \(P(A|B) = \frac{P(A \cap B)}{P(B)}\).
Step 3: Detailed Explanation:
Let's use the strict formula to be certain.
The sample space for rolling a die twice has \(6 \times 6 = 36\) outcomes.
Event A (1 on first roll): \(\{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)\}\).
\(P(A) = \frac{6}{36} = \frac{1}{6}\).
Event B (4 on second roll): \(\{(1,4), (2,4), (3,4), (4,4), (5,4), (6,4)\}\).
\(P(B) = \frac{6}{36} = \frac{1}{6}\).
Event \(A \cap B\) (1 on first roll AND 4 on second roll): \(\{(1,4)\}\).
\(P(A \cap B) = \frac{1}{36}\).
Now calculate \(P(A|B)\):
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1/36}{1/6} = \frac{1}{36} \times \frac{6}{1} = \frac{1}{6} \] Notice this is exactly equal to \(P(A)\) because the events are independent.
Step 4: Final Answer:
The probability is \(\frac{1}{6}\).