Question

Let \(X\sim B(n,p)\). If \(E(X) = 2\) and \(\text{Var}(X) = 1\), then the probability of getting at most one success is .....

• A. \(0.0625\)
• B. \(0.2500\)
• C. \(0.3125\)
• D. \(0.6875\)

Hint:

Use $np=2$ and $npq=1$ to find $n$ and $p$.

Answer 1

The Correct Option is C

Step 1: Find \(p\) and \(n\)
Mean \(np=2\) and variance \(npq=1\). Dividing, \(q=\frac12\), so \(p=\frac12\) and \(n=4\).

Step 2: Compute
\[ P(X\le1)=\binom40\left(\tfrac12\right)^4+\binom41\left(\tfrac12\right)^4=\frac{1+4}{16}=\frac{5}{16}=0.3125 \]

Final Answer:
The probability is \(0.3125\), option (C). \[ \boxed{\text{(C) }0.3125} \]