Question

An experiment is performed by rolling an unbiased die once. Events A and B are defined as follows:
Event A: The outcome of the rolled die is a prime number.
Event B: The outcome of the rolled die is an odd number less than 4.
The conditional probability of A given B, denoted by P(A|B), is ______ (rounded off to one decimal place).

Hint:

List the outcomes in event B, then find what fraction of those outcomes are also prime.

Answer 1

Correct Answer: 0.5

Step 1: Define the sample space.
Rolling an unbiased die once gives the sample space \(S = \{1,2,3,4,5,6\}\), each outcome equally likely with probability \(\dfrac{1}{6}\).

Step 2: Identify Event A.
A: outcome is a prime number. Primes in \(S\) are \(2, 3, 5\).
\(A = \{2,3,5\}\), so \(P(A) = \dfrac{3}{6} = 0.5\).

Step 3: Identify Event B.
B: outcome is an odd number less than 4. Odd numbers in \(S\) are \(1,3,5\); of these, those less than 4 are \(1\) and \(3\).
\(B = \{1,3\}\), so \(P(B) = \dfrac{2}{6} = \dfrac{1}{3}\).

Step 4: Find the intersection \(A \cap B\).
Compare the two sets: \(A=\{2,3,5\}\), \(B=\{1,3\}\). The only common element is \(3\).
\(A \cap B = \{3\}\), so \(P(A \cap B) = \dfrac{1}{6}\).

Step 5: Apply the conditional probability formula.
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1/6}{2/6} = \frac{1}{2} = 0.5 \]

\[ \boxed{P(A|B) = 0.5} \]