Q266 / 270 · NDA 2025 (Phase 2) · Mathematics
Let $X$ be a random variable following binomial distribution with parameters $n = 5$ and $p = k$. Further, $P(X = 1) = 0.4096$ and $P(X = 2) = 0.2048$. What is the value of $k$?
- Correct answerA$0.2$
- B$0.25$
- C$0.3$
- D$0.35$
Solution
Given $X$ follows a binomial distribution with parameters $n = 5$ and $p = k$.
The probability mass function is given by $P(X = x) = ^{n}C_{x} p^{x} (1-p)^{n-x}$.
For $x = 1$:
$P(X = 1) = ^{5}C_{1} k^{1} (1-k)^{4} = 5k(1-k)^{4} = 0.4096$
For $x = 2$:
$P(X = 2) = ^{5}C_{2} k^{2} (1-k)^{3} = 10k^{2}(1-k)^{3} = 0.2048$
Dividing $P(X = 1)$ by $P(X = 2)$, we get:
$\dfrac{P(X = 1)}{P(X = 2)} = \dfrac{5k(1-k)^{4}}{10k^{2}(1-k)^{3}}$
$\dfrac{0.4096}{0.2048} = \dfrac{1-k}{2k}$
$2 = \dfrac{1-k}{2k}$
$4k = 1 - k$
$5k = 1$
$k = 0.2$
The probability mass function is given by $P(X = x) = ^{n}C_{x} p^{x} (1-p)^{n-x}$.
For $x = 1$:
$P(X = 1) = ^{5}C_{1} k^{1} (1-k)^{4} = 5k(1-k)^{4} = 0.4096$
For $x = 2$:
$P(X = 2) = ^{5}C_{2} k^{2} (1-k)^{3} = 10k^{2}(1-k)^{3} = 0.2048$
Dividing $P(X = 1)$ by $P(X = 2)$, we get:
$\dfrac{P(X = 1)}{P(X = 2)} = \dfrac{5k(1-k)^{4}}{10k^{2}(1-k)^{3}}$
$\dfrac{0.4096}{0.2048} = \dfrac{1-k}{2k}$
$2 = \dfrac{1-k}{2k}$
$4k = 1 - k$
$5k = 1$
$k = 0.2$
Hence, the correct answer is option: 'A'.