Question

If \( X \sim B(9, p) \) is a binomial variate satisfying the equation \( P(X = 3) = P(X = 6) \), then \( P(X < 3) = \)?

• A. $ \frac{23}{256} $
• B. $ \frac{65}{256} $
• C. $ \frac{5}{256} $
• D. $ \frac{45}{512} $

Hint:

When solving problems involving binomial distributions, use the symmetry properties of the binomial coefficients and the given conditions to simplify the calculations. For cumulative probabilities, sum the individual probabilities up to the desired value.

Answer 1

The Correct Option is A

Step 1: Use the given condition $ P(X = 3) = P(X = 6) $.
The probability mass function (PMF) of a binomial distribution is: $$ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}, $$ where $ n = 9 $ and $ k = 3, 6 $. Given $ P(X = 3) = P(X = 6) $, we have: $$ \binom{9}{3} p^3 (1-p)^6 = \binom{9}{6} p^6 (1-p)^3. $$ Since $ \binom{9}{3} = \binom{9}{6} $, this simplifies to: $$ p^3 (1-p)^6 = p^6 (1-p)^3. $$ Divide both sides by $ p^3 (1-p)^3 $ (assuming $ p \neq 0 $ and $ p \neq 1 $): $$ (1-p)^3 = p^3. $$ Take the cube root: $$ 1 - p = p \quad \Rightarrow \quad p = \frac{1}{2}. $$ Step 2: Compute $ P(X<3) $.
Now that $ p = \frac{1}{2} $, the PMF becomes:
$$ P(X = k) = \binom{9}{k} \left(\frac{1}{2}\right)^k \left(\frac{1}{2}\right)^{9-k} = \binom{9}{k} \left(\frac{1}{2}\right)^9. $$ We need $ P(X<3) = P(X = 0) + P(X = 1) + P(X = 2) $: $$ P(X<3) = \sum_{k=0}^{2} P(X = k) = \sum_{k=0}^{2} \binom{9}{k} \left(\frac{1}{2}\right)^9. $$ Compute each term: $$ P(X = 0) = \binom{9}{0} \left(\frac{1}{2}\right)^9 = 1 \cdot \frac{1}{512} = \frac{1}{512}, $$ $$ P(X = 1) = \binom{9}{1} \left(\frac{1}{2}\right)^9 = 9 \cdot \frac{1}{512} = \frac{9}{512}, $$ $$ P(X = 2) = \binom{9}{2} \left(\frac{1}{2}\right)^9 = 36 \cdot \frac{1}{512} = \frac{36}{512}. $$ Sum these probabilities:
$$ P(X<3) = \frac{1}{512} + \frac{9}{512} + \frac{36}{512} = \frac{1 + 9 + 36}{512} = \frac{46}{512} = \frac{23}{256}. $$ Step 3: Final Answer.
$$ \boxed{\frac{23}{256}} $$