Question

If \(X\sim B(n,p)\), then the value of \(\frac{P(X = k)}{P(X = k-1)}\) is

• A. \((\frac{n-k}{k})\frac{p}{q}\)
• B. \((\frac{n-k+1}{k})\frac{p}{q}\)
• C. \((\frac{n-k}{k-1})\frac{p}{q}\)
• D. \((\frac{n-k+1}{k-1})\frac{p}{q}\)

Hint:

Write both binomial probabilities and divide; the powers of p and q partly cancel.

Answer 1

The Correct Option is B

Step 1: Test with a small case:
Take $n=2$, $k=1$: $P(1)=2pq$, $P(0)=q^2$. The ratio is $\dfrac{2p}{q}$.

Step 2: Check each option with n = 2, k = 1:
(A): $\dfrac{1}{1}\cdot\dfrac pq=\dfrac pq$, wrong. (B): $\dfrac{2}{1}\cdot\dfrac pq=\dfrac{2p}{q}$, correct. (C) and (D) have $k-1=0$ in the denominator, which is undefined.

Step 3: Pick:
Option B.

Final Answer:
Dividing the two binomial terms gives (n - k + 1)/k times p/q. \[ \boxed{\text{(B) }\left(\dfrac{n-k+1}{k}\right)\dfrac pq} \]