Question

Let \(X\sim B(6,\frac{1}{2})\). Then the maximum probability occurs at

• A. \(X = 1\)
• B. \(X = 2\)
• C. \(X = 3\)
• D. \(X = 4\)

Hint:

Compare the binomial probabilities, or use the mode formula for a binomial distribution.

Answer 1

The Correct Option is C

Step 1: Use the mode rule.
The mode of $B(n, p)$ is the integer part of $(n+1)p$ when this is not an integer.

Step 2: Compute.
$(n + 1)p = 7 \times \dfrac{1}{2} = 3.5$, so the mode is 3.

Step 3: Check neighbours.
$P(3) = 20/64$ is more than $P(2) = P(4) = 15/64$.

Final Answer:
Option (C). \[ \boxed{X = 3} \]