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: Write both probabilities:
\(P(X=k)=\binom nk p^kq^{n-k}\) and \(P(X=k-1)=\binom n{k-1}p^{k-1}q^{n-k+1}\).

Step 2: Divide:
\(\dfrac{P(X=k)}{P(X=k-1)}=\dfrac{\binom nk}{\binom n{k-1}}\cdot\dfrac pq\).

Step 3: Use the ratio of binomial coefficients:
\(\dfrac{\binom nk}{\binom n{k-1}}=\dfrac{(k-1)!(n-k+1)!}{k!(n-k)!}=\dfrac{n-k+1}{k}\).

Step 4: Result:
\[ \dfrac{P(X=k)}{P(X=k-1)}=\dfrac{n-k+1}{k}\cdot\dfrac pq \]
Option B.

Final Answer:
The ratio is ((n - k + 1)/k)(p/q). \[ \boxed{\text{(B) }\left(\dfrac{n-k+1}{k}\right)\dfrac pq} \]