Question

A random variable \(X\sim B(n,p)\) follows a binomial distribution with \(n = 6\). If \(9P(X = 4) = P(X = 2)\), then the probability of success \(p\) is..

• A. \(0.125\)
• B. \(0.75\)
• C. \(0.25\)
• D. \(0.375\)

Hint:

Write both probabilities with C(6,r) p^r q^(6-r) and compare.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For \(X \sim B(6, p)\), \(P(X = r) = \binom6r p^r q^{6-r}\) with \(q = 1 - p\).

Step 2: Write the condition:
\(9P(X = 4) = P(X = 2)\) gives
\[ 9\binom64 p^4q^2 = \binom62 p^2q^4 \]
Since \(\binom64 = \binom62 = 15\), they cancel: \(9p^4q^2 = p^2q^4\).

Step 3: Simplify:
Divide by \(p^2q^2\): \(9p^2 = q^2\), so \(3p = q = 1 - p\) (taking positive roots).
Then \(4p = 1\), so \(p = \frac14 = 0.25\).

Step 4: Why the other options are wrong.
\(p = 0.125\) gives \(q = 0.875 \ne 3p\). \(p = 0.75\) gives \(q = 0.25\), so \(q = \frac{p}{3}\), the reverse ratio. \(p = 0.375\) gives \(q = 0.625 \ne 1.125\).

Final Answer:
The probability of success is \(0.25\), option (C). \[ \boxed{0.25} \]