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
$1 / 6$
B
$3 / 64$
C
$9 / 64$
D
$2 / 3$

Solution:

We have,

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

$∴ q = 1 - p = 1 - \frac{1}{4} = \frac{3}{4}$

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

$\begin{array}{l} P (X = r) =^{3} C_{r} {(\frac{1}{4})}^{r} {(\frac{3}{4})}^{3 - r} \\ \Rightarrow P (X = 2) =^{3} C_{2} {(\frac{1}{4})}^{2} (\frac{3}{4}) = \frac{9}{64} \end{array}$

Solution:

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