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 \( \frac{200}{243} \).