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A box has \( 100 \) pens of which \( 10 \) are defective. The probability that out of a sample of \( 5 \) pens
drawn one by one with replacement and at most one is defective is
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drawn one by one with replacement and at most one is defective is
Solution:
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Verified Answer
The correct answer is:
\( \left(\frac{9}{10}\right)^{5}+\frac{1}{2}\left(\frac{9}{10}\right)^{4} \)
Given that, probability of defective pens is
\[
P=\frac{10}{100}=\frac{1}{10}
\]
So, probability of non-defective pens is
\[
\begin{array}{l}
Q=1-P=\frac{9}{10} \\
\text { Since, } n=5, x \leq 1 \text {. So, } \\
P(x \leq 1)=P(x=0)+P(x=1) \\
={ }^{5} C_{0}\left(\frac{1}{10}\right)^{0}\left(\frac{9}{10}\right)^{5}+{ }^{5} C_{1}\left(\frac{1}{10}\right)^{1}\left(\frac{9}{10}\right)^{4} \\
=\left(\frac{9}{10}\right)^{5}+\frac{1}{2}\left(\frac{9}{10}\right)^{4}
\end{array}
\]
\[
P=\frac{10}{100}=\frac{1}{10}
\]
So, probability of non-defective pens is
\[
\begin{array}{l}
Q=1-P=\frac{9}{10} \\
\text { Since, } n=5, x \leq 1 \text {. So, } \\
P(x \leq 1)=P(x=0)+P(x=1) \\
={ }^{5} C_{0}\left(\frac{1}{10}\right)^{0}\left(\frac{9}{10}\right)^{5}+{ }^{5} C_{1}\left(\frac{1}{10}\right)^{1}\left(\frac{9}{10}\right)^{4} \\
=\left(\frac{9}{10}\right)^{5}+\frac{1}{2}\left(\frac{9}{10}\right)^{4}
\end{array}
\]
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