Question

If \( X \sim B(5, p) \) is a binomial variate such that \( p(X = 3) = p(X = 4) \), then \( P(|X - 3| < 2) = \dots \)

• A. \( \frac{242}{243} \)
• B. \( \frac{201}{243} \)
• C. \( \frac{200}{243} \)
• D. \( \frac{121}{243} \)

Hint:

When solving binomial distribution problems, use the probability mass function to relate different outcomes and solve for unknown probabilities.

Answer 1

The Correct Option is C

Let \( X \sim B(5, p) \) be a binomial random variable. The probability mass function of a binomial random variable is given by: \[ P(X = k) = \binom{5}{k} p^k (1-p)^{5-k}, \quad k = 0, 1, 2, \dots, 5. \] We are given that \( P(X = 3) = P(X = 4) \). Hence, we have: \[ \binom{5}{3} p^3 (1-p)^2 = \binom{5}{4} p^4 (1-p). \] Simplifying: \[ 10 p^3 (1-p)^2 = 5 p^4 (1-p), \] \[ 2 p^3 (1-p) = p^4, \] \[ 2 (1-p) = p, \] \[ 2 - 2p = p, \] \[ 3p = 2, \] \[ p = \frac{2}{3}. \] Now, we need to calculate \( P(|X - 3| < 2) \), which is equivalent to \( P(1 \leq X \leq 5) \). Using the binomial distribution: \[ P(1 \leq X \leq 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5). \] Using the values of \( p \) and the binomial probabilities: \[ P(1 \leq X \leq 5) = \frac{200}{243}. \] Thus, the correct answer is \( \frac{200}{243} \).

Answer 2

Given \( X \sim B(5, p) \) and \( p(X = 3) = p(X = 4) \), find \( P(|X - 3| < 2) \).

Step 1: Use the binomial pmf:
\[ P(X = k) = \binom{5}{k} p^k (1-p)^{5-k} \] Given:
\[ P(X = 3) = P(X = 4) \] \[ \binom{5}{3} p^3 (1-p)^2 = \binom{5}{4} p^4 (1-p)^1 \] \[ 10 p^3 (1-p)^2 = 5 p^4 (1-p) \] Divide both sides by \( 5 p^3 (1-p) \):
\[ 2 (1-p) = p \implies 2 - 2p = p \implies 2 = 3p \implies p = \frac{2}{3} \]

Step 2: Calculate \( P(|X - 3| < 2) \), i.e., \( P(2 \leq X \leq 4) \):
\[ P(2 \leq X \leq 4) = P(X=2) + P(X=3) + P(X=4) \]

Step 3: Calculate each term with \( p = \frac{2}{3} \), \( q = 1-p = \frac{1}{3} \):
\[ P(X=2) = \binom{5}{2} \left(\frac{2}{3}\right)^2 \left(\frac{1}{3}\right)^3 = 10 \times \frac{4}{9} \times \frac{1}{27} = \frac{40}{243} \] \[ P(X=3) = \binom{5}{3} \left(\frac{2}{3}\right)^3 \left(\frac{1}{3}\right)^2 = 10 \times \frac{8}{27} \times \frac{1}{9} = \frac{80}{243} \] \[ P(X=4) = \binom{5}{4} \left(\frac{2}{3}\right)^4 \left(\frac{1}{3}\right)^1 = 5 \times \frac{16}{81} \times \frac{1}{3} = \frac{80}{243} \]

Step 4: Sum:
\[ P(2 \leq X \leq 4) = \frac{40}{243} + \frac{80}{243} + \frac{80}{243} = \frac{200}{243} \]

Therefore,
\[ \boxed{\frac{200}{243}} \]