Q96 / 150 · MHT CET 2026 (11 April Shift 1) · Mathematics
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$0.2500$
- Correct answerC$0.3125$
- D$0.6875$
Solution
For a binomial distribution $X \sim B(n, p)$, the mean and variance are given by:
$E(X) = np = 2$
$\text{Var}(X) = npq = 1$
Dividing the variance by the mean:
$q = \dfrac{npq}{np} = \dfrac{1}{2}$
Since $p + q = 1$, we get $p = \dfrac{1}{2}$.
Substituting $p$ in $np = 2$:
$n \left(\dfrac{1}{2}\right) = 2 \Rightarrow n = 4$
The probability of getting at most one success is:
$P(X \le 1) = P(X = 0) + P(X = 1)$
$P(X \le 1) = ^{4}C_{0} \left(\dfrac{1}{2}\right)^0 \left(\dfrac{1}{2}\right)^4 + ^{4}C_{1} \left(\dfrac{1}{2}\right)^1 \left(\dfrac{1}{2}\right)^3$
$P(X \le 1) = \dfrac{1}{16} + \dfrac{4}{16} = \dfrac{5}{16}$
$P(X \le 1) = 0.3125$
Answer: $0.3125$
$E(X) = np = 2$
$\text{Var}(X) = npq = 1$
Dividing the variance by the mean:
$q = \dfrac{npq}{np} = \dfrac{1}{2}$
Since $p + q = 1$, we get $p = \dfrac{1}{2}$.
Substituting $p$ in $np = 2$:
$n \left(\dfrac{1}{2}\right) = 2 \Rightarrow n = 4$
The probability of getting at most one success is:
$P(X \le 1) = P(X = 0) + P(X = 1)$
$P(X \le 1) = ^{4}C_{0} \left(\dfrac{1}{2}\right)^0 \left(\dfrac{1}{2}\right)^4 + ^{4}C_{1} \left(\dfrac{1}{2}\right)^1 \left(\dfrac{1}{2}\right)^3$
$P(X \le 1) = \dfrac{1}{16} + \dfrac{4}{16} = \dfrac{5}{16}$
$P(X \le 1) = 0.3125$
Answer: $0.3125$
Hence, the correct answer is option: 'C'.