Question

If \( X \sim B(35, p) \) such that \( 7P(X = 0) = P(X = 1) \), then the value of

\[ \frac{P(X = 15)}{P(X = 20)} \] 

is ______.

• A. (\frac{3125}{7776})
• B. (3125)
• C. (7776)
• D. [suspicious link removed]

Hint:

In Binomial distribution, the ratio $\frac{P(X=k)}{P(X=k-1)}$ helps in finding $p$ quickly.

Answer 1

The Correct Option is D

Step 1: Find $p$
$7 \cdot {}^35C_0 \cdot q^{35} = {}^35C_1 \cdot p^1 \cdot q^{34}$.
$7 q^{35} = 35 p q^{34} \implies 7q = 35p \implies q = 5p$.
Since $p+q=1$, $p+5p=1 \implies 6p=1 \implies p=1/6, q=5/6$.

Step 2: Evaluate the ratio
$\frac{{}^35C_{15} \cdot p^{15} q^{20}}{{}^35C_{20} \cdot p^{20} q^{15}}$.
Since ${}^nC_r = {}^nC_{n-r}$, ${}^35C_{15} = {}^35C_{20}$. The ratio simplifies to $\frac{q^5}{p^5}$.

Step 3: Calculation
Ratio $= \left(\frac{q}{p}\right)^5 = \left(\frac{5/6}{1/6}\right)^5 = 5^5 = 3125$.
*(Re-checking option mapping: if ratio is P(20)/P(15), it is 1/3125; based on source data, value is 3125).*
Final Answer: (B)