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# The probability of a man hitting a target is $\large\frac{3}{4}$. He tries 5 times. The probability that the target will be hit at least 3 times is

$\begin {array} {1 1} (A)\;\large\frac{291}{364} & \quad (B)\;\large\frac{371}{464} \\ (C)\;\large\frac{471}{502} & \quad (D)\;\large\frac{459}{512} \end {array}$

Can you answer this question?

## 1 Answer

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Using binomial distribution we have to find $P(3) +P(4)+P(5)$
This probability is
$5C_3 \bigg( \large\frac{3}{4} \bigg)^3\bigg( \large\frac{1}{4} \bigg)^2+5C_4\bigg( \large\frac{3}{4} \bigg)^4\bigg( \large\frac{1}{4} \bigg)^1+5C_5\bigg( \large\frac{3}{4} \bigg)^5\bigg( \large\frac{1}{4} \bigg)^0$
$= \large\frac{270+405+243}{45} = \large\frac{918}{1024}$
$= \large\frac{459}{512}$
Ans : (D)
answered Jan 3, 2014

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