Question

Two dice are rolled. If A is the event that sum of the numbers is 4 and B is the event that at least one of the dice shows a 3, then P(A|B) is equal to

• A. \(\frac{3}{11}\)
• B. \(\frac{2}{11}\)
• C. \(\frac{1}{4}\)
• D. \(\frac{1}{6}\)
• E. \(\frac{1}{11}\)

Answer 1

The Correct Option is B

Let the sample space \( S \) represent all possible outcomes when two dice are rolled. The total number of possible outcomes is \( 6 \times 6 = 36 \). Hence, \( |S| = 36 \).

Event A: The sum of the numbers on the two dice is 4. The favorable outcomes are:

• (1, 3)
• (2, 2)
• (3, 1)

So, \( |A| = 3 \).

Event B: At least one of the dice shows a 3. The favorable outcomes are:

• (3, 1)
• (3, 2)
• (3, 3)
• (3, 4)
• (3, 5)
• (3, 6)
• (1, 3)
• (2, 3)
• (4, 3)
• (5, 3)
• (6, 3)

So, \( |B| = 11 \).

Event A ∩ B: The favorable outcomes where the sum is 4 and at least one die shows a 3 are:

• (1, 3)
• (3, 1)

So, \( |A \cap B| = 2 \).

Now, we use the conditional probability formula:

\[ P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{|A \cap B|}{|B|} \] Substitute the values: \[ P(A|B) = \frac{2}{11} \]

Answer: \( \frac{2}{11} \)