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:
The rolling of a die twice involves independent events. The result of the first roll does not affect the result of the second roll. For independent events, $P(A|B) = P(A)$.

Step 2: Detailed Explanation:
Total outcomes when a die is rolled twice = $6 \times 6 = 36$.
Event $A$: getting 1 in the first roll.
$A = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)\}$.
$P(A) = 6/36 = 1/6$.
Event $B$: getting 4 in the second roll.
$B = \{(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4)\}$.
$P(B) = 6/36 = 1/6$.
Intersection $A \cap B$: getting 1 in first roll AND 4 in second roll.
$A \cap B = \{(1, 4)\}$.
$P(A \cap B) = 1/36$.
By conditional probability formula:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1/36}{1/6} = \frac{6}{36} = \frac{1}{6} \]
Since $P(A|B) = P(A)$, the events are independent.

Step 3: Final Answer:
$P(A|B) = 1/6$.