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a box contains 20 percent defective bulbs five bul
Question:
A box contains 20 percent defective bulbs. Five bulbs are randomly chosen. Find the probability that exactly 3 are defective.
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Use binomial probability distribution for repeated trials with success/failure outcomes.
AP EAMCET - 2024
AP EAMCET
Updated On:
Mar 19, 2025
\( \frac{32}{625} \)
\( \frac{32}{125} \)
\( \frac{16}{625} \)
\( \frac{16}{125} \)
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The Correct Option is
A
Solution and Explanation
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 \( \frac{32}{625} \).
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