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# A fair coin is tossed 10 times. The probability that the coin shows head exactly 4 times is

$\begin {array} {1 1} (A)\;\large\frac{105}{512} & \quad (B)\;\large\frac{103}{512} \\ (C)\;\large\frac{101}{512} & \quad (D)\;None\: of \: these \end {array}$

Here $n = 10$
$\Rightarrow P = \large\frac{1}{2}$
$Q = \large\frac{1}{2}$
Hence $P(4)=10C_4 \bigg( \large\frac{1}{2} \bigg)^4 \bigg( \large\frac{1}{2} \bigg)^6$
$= \large\frac{210}{1024}$
$= \large\frac{105}{512}$