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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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The correct answer is:
0.8192
We have,
$n=4, p=\frac{20}{100}=\frac{1}{5}, q=1-p=1-\frac{1}{5}=\frac{4}{5}$
$\begin{aligned} & \therefore \text { Required probability }=P(x < 2) \\ & =P(x=0)+P(x=1) \\ & ={ }^4 C_0\left(\frac{1}{5}\right)^0\left(\frac{4}{5}\right)^4+{ }^4 C_1\left(\frac{1}{5}\right)^1\left(\frac{4}{5}\right)^3 \\ & =\left(\frac{4}{5}\right)^4+\left(\frac{4}{5}\right)^4=2\left(\frac{4}{5}\right)^4=0.8192 \\ & \end{aligned}$
$n=4, p=\frac{20}{100}=\frac{1}{5}, q=1-p=1-\frac{1}{5}=\frac{4}{5}$
$\begin{aligned} & \therefore \text { Required probability }=P(x < 2) \\ & =P(x=0)+P(x=1) \\ & ={ }^4 C_0\left(\frac{1}{5}\right)^0\left(\frac{4}{5}\right)^4+{ }^4 C_1\left(\frac{1}{5}\right)^1\left(\frac{4}{5}\right)^3 \\ & =\left(\frac{4}{5}\right)^4+\left(\frac{4}{5}\right)^4=2\left(\frac{4}{5}\right)^4=0.8192 \\ & \end{aligned}$
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