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A box contains tickets numbered from 1 to 20. If 3 tickets are drawn one by one with replacement, then the probability of getting prime number exactly 2 times is

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$\frac{4}{125}$

$\frac{1}{125}$

$\frac{12}{125}$

$\frac{36}{125}$

answer is D.

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Detailed Solution

p = prime $= \frac{8}{20} = \frac{2}{5} \Rightarrow q = \frac{3}{5}, n = 3$
$∴ p (X = 2) =^{3} c_{2} {(\frac{2}{5})}^{2} {(\frac{3}{5})}^{1} = \frac{36}{125}$

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