Q100 / 150 · MHT CET 2026 (11 April Shift 2) · Mathematics
If $X \sim B(n, p)$, then the value of $\dfrac{P(X = k)}{P(X = k - 1)}$ is
- A$\left(\dfrac{n - k}{k}\right)\dfrac{p}{q}$
- Correct answerB$\left(\dfrac{n - k + 1}{k}\right)\dfrac{p}{q}$
- C$\left(\dfrac{n - k}{k - 1}\right)\dfrac{p}{q}$
- D$\left(\dfrac{n - k + 1}{k - 1}\right)\dfrac{p}{q}$
Solution
For a binomial distribution $X \sim B(n, p)$, the probability of $k$ successes is given by $P(X = k) = ^{n}C_{k} p^k q^{n-k}$, where $q = 1 - p$.
Substituting $k$ and $k-1$, we get:
$P(X = k) = ^{n}C_{k} p^k q^{n-k}$
$P(X = k - 1) = ^{n}C_{k-1} p^{k-1} q^{n-k+1}$
Taking the ratio of the two probabilities:
$\dfrac{P(X = k)}{P(X = k - 1)} = \dfrac{^{n}C_{k} p^k q^{n-k}}{^{n}C_{k-1} p^{k-1} q^{n-k+1}}$
$\dfrac{P(X = k)}{P(X = k - 1)} = \dfrac{^{n}C_{k}}{^{n}C_{k-1}} \cdot \dfrac{p}{q}$
Using the property of combinations $\dfrac{^{n}C_{r}}{^{n}C_{r-1}} = \dfrac{n - r + 1}{r}$:
$\dfrac{P(X = k)}{P(X = k - 1)} = \left(\dfrac{n - k + 1}{k}\right)\dfrac{p}{q}$
Answer: $\left(\dfrac{n - k + 1}{k}\right)\dfrac{p}{q}$
Substituting $k$ and $k-1$, we get:
$P(X = k) = ^{n}C_{k} p^k q^{n-k}$
$P(X = k - 1) = ^{n}C_{k-1} p^{k-1} q^{n-k+1}$
Taking the ratio of the two probabilities:
$\dfrac{P(X = k)}{P(X = k - 1)} = \dfrac{^{n}C_{k} p^k q^{n-k}}{^{n}C_{k-1} p^{k-1} q^{n-k+1}}$
$\dfrac{P(X = k)}{P(X = k - 1)} = \dfrac{^{n}C_{k}}{^{n}C_{k-1}} \cdot \dfrac{p}{q}$
Using the property of combinations $\dfrac{^{n}C_{r}}{^{n}C_{r-1}} = \dfrac{n - r + 1}{r}$:
$\dfrac{P(X = k)}{P(X = k - 1)} = \left(\dfrac{n - k + 1}{k}\right)\dfrac{p}{q}$
Answer: $\left(\dfrac{n - k + 1}{k}\right)\dfrac{p}{q}$
Hence, the correct answer is option: 'B'.