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If 20% of the bolts produced by a machine are defective then the probability that out of 4 bolts chosen at random, less than 2 bolts will be defective is

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a

0.4096

b

0.1024

c

0.8192

d

0.2048

answer is C.

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Detailed Solution

$A = 4, P = \frac{20}{100} = \frac{1}{5}, q = 1 - p = 1 - \frac{1}{5} = \frac{4}{5}$
Required probability $= p (x < 2)$
$= p (x = 0) + p (x - 1)$
$\begin{array}{l} = {}^{4}C_{0} {(\frac{1}{5})}^{0} {(\frac{4}{5})}^{4} + {}^{4}C_{1} {(\frac{1}{5})}^{1} {(\frac{4}{5})}^{3} \\ = {(\frac{4}{5})}^{4} + {(\frac{4}{5})}^{4} = 2 {(\frac{4}{5})}^{4} = 0.8192 \end{array}$

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