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A lot of 100 bulbs contains 10 defective bulbs. Five bulbs are selected at random from the lot and are sent to retail store. Then the probability that the store will receive at most one defective bulb is
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Verified Answer
The correct answer is:
$\frac{7}{5}\left(\frac{9}{10}\right)^4$
Let $\mathrm{X}$ denote the number of defective bulbs. $\mathrm{p}=$ Probability that a bulb is defective
$\begin{aligned}
& =\frac{10}{100}=\frac{1}{10} \\
& q=1-p=1-\frac{1}{10}=\frac{9}{10} \\
& P(X=r)={ }^5 C_r\left(\frac{1}{10}\right)^r\left(\frac{9}{10}\right)^{5-r}, r=0,1, \ldots, 5
\end{aligned}$
$\begin{aligned}
\therefore \quad & P(X \leq 1) \\
& =P(X=0)+P(X=1) \\
& ={ }^5 C_0\left(\frac{1}{10}\right)^0\left(\frac{9}{10}\right)^5+{ }^5 C_1\left(\frac{1}{10}\right)^1\left(\frac{9}{10}\right)^4 \\
& =\left(\frac{9}{10}\right)^5+5 \times \frac{1}{10} \times\left(\frac{9}{10}\right)^4 \\
& =\frac{7}{5}\left(\frac{9}{10}\right)^4
\end{aligned}$
$\begin{aligned}
& =\frac{10}{100}=\frac{1}{10} \\
& q=1-p=1-\frac{1}{10}=\frac{9}{10} \\
& P(X=r)={ }^5 C_r\left(\frac{1}{10}\right)^r\left(\frac{9}{10}\right)^{5-r}, r=0,1, \ldots, 5
\end{aligned}$
$\begin{aligned}
\therefore \quad & P(X \leq 1) \\
& =P(X=0)+P(X=1) \\
& ={ }^5 C_0\left(\frac{1}{10}\right)^0\left(\frac{9}{10}\right)^5+{ }^5 C_1\left(\frac{1}{10}\right)^1\left(\frac{9}{10}\right)^4 \\
& =\left(\frac{9}{10}\right)^5+5 \times \frac{1}{10} \times\left(\frac{9}{10}\right)^4 \\
& =\frac{7}{5}\left(\frac{9}{10}\right)^4
\end{aligned}$
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