Question

A box contains 20 percent defective bulbs. Five bulbs are randomly chosen. Find the probability that exactly 3 are defective.

• A. \( \frac{32}{625} \)
• B. \( \frac{32}{125} \)
• C. \( \frac{16}{625} \)
• D. \( \frac{16}{125} \)

Hint:

Use binomial probability distribution for repeated trials with success/failure outcomes.

Answer 1

The Correct Option is A

Step 1: Binomial Probability Formula
\[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \] Given: \( n = 5, k = 3, p = 0.2 \), \[ P(X = 3) = \binom{5}{3} (0.2)^3 (0.8)^2 \] Step 2: Computation
\[ = 10 \times (0.008) \times (0.64) \] \[ = 10 \times 0.00512 = 0.0512 \] \[ = \frac{32}{625} \] Thus, the correct answer is \( \frac{32}{625} \).