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: Understanding the Concept:
For \(X \sim B(n, p)\), \(P(X = r) = {}^nC_r\,p^r q^{n-r}\). Here \(p = q = \dfrac{1}{2}\), so \(P(X = r) = {}^6C_r\left(\dfrac{1}{2}\right)^6\). The probability is largest where \({}^6C_r\) is largest.

Step 2: Compare the coefficients.
\({}^6C_0 = 1\), \({}^6C_1 = 6\), \({}^6C_2 = 15\), \({}^6C_3 = 20\), \({}^6C_4 = 15\), \({}^6C_5 = 6\), \({}^6C_6 = 1\).

Step 3: Pick the maximum.
The largest is 20 at \(r = 3\). So the maximum probability \(\dfrac{20}{64}\) occurs at \(X = 3\).

Final Answer:
The maximum probability occurs at \(X = 3\), option (C). \[ \boxed{X = 3} \]