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The probability that a certain kind of component will survive a given shock test is $\frac{3}{4}$. The probability that exactly 2 of the next 4 components tested survive is
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The correct answer is:
$\frac{27}{128}$
The probability that a component survives is $\mathrm{p}=\frac{3}{4}$. Then $\mathrm{q}=1-\mathrm{p}=1-\frac{3}{4}=\frac{1}{4}$
$[\because \mathrm{p}+\mathrm{q}=1]$
$\mathrm{n}$ takes the value 4 and $\mathrm{r}=2$. Hence the required probability is
${ }^{n} C_{r} p^{r} q^{n-r}={ }^{4} C_{2}\left(\frac{3}{4}\right)^{2}\left(\frac{1}{4}\right)^{2}$
$=6 \times \frac{3 \times 3 \times 1}{4 \times 4 \times 4 \times 4}=\frac{27}{128}$
$[\because \mathrm{p}+\mathrm{q}=1]$
$\mathrm{n}$ takes the value 4 and $\mathrm{r}=2$. Hence the required probability is
${ }^{n} C_{r} p^{r} q^{n-r}={ }^{4} C_{2}\left(\frac{3}{4}\right)^{2}\left(\frac{1}{4}\right)^{2}$
$=6 \times \frac{3 \times 3 \times 1}{4 \times 4 \times 4 \times 4}=\frac{27}{128}$
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