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# A coin is tossed 7 times. The probability that he wins the toss on more occasions is

$\begin {array} {1 1} (A)\;\large\frac{1}{4} & \quad (B)\;\large\frac{5}{8} \\ (C)\;\large\frac{1}{2} & \quad (D)\;None\: of \: these \end {array}$

The man has to win atleast 4 times.
$\therefore$ The required probability
$= 7C_4 \bigg( \large\frac{1}{2} \bigg)^4 \bigg( \large\frac{1}{2} \bigg)^3 + 7C_5 \bigg( \large\frac{1}{2} \bigg)^5 \bigg( \large\frac{1}{2} \bigg)^2 +7C_6 \bigg( \large\frac{1}{2} \bigg)^6 \large\frac{1}{2} + 7C_7 \bigg( \large\frac{1}{2} \bigg)^7$
$= ( 7C_4+7C_5+7C_6+7C7 ). \large\frac{1}{2^7}$
$= \large\frac{6^4}{2^7}$
$= \large\frac{1}{2}$
Ans : (C)