A carton contains 20 bulbs, 5 of which are defective. The probability that, if a sample of 3 bulbs is chosen at random from the carton 2 will be defective, is

# A carton contains 20 bulbs, 5 of which are defective. The probability that, if a sample of 3 bulbs is chosen at random from the carton 2 will be defective, is

1. A

$1/6$

2. B

$3/64$

3. C

$9/64$

4. D

$2/3$

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### Solution:

We have,

Probability that a bulb is defective $=\frac{5}{20}=\frac{1}{4}$

Let $X$ denote the number of defective bulbs in a sample of 3 bulbs. Then, X is a binomial variate with parameter and $p=\frac{1}{4}$ such that

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