A fair coin is tossed 10 times. The probability of getting as many heads in the first 5 tosses as in the last 5 tosses is

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A fair coin is tossed 10 times. The probability of getting as many heads in the first 5 tosses as in the last 5 tosses is

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$\frac{1}{2^{10}}$

$\frac{^{10} C_{5}}{2^{10}}$

$\frac{1}{2^{9}}$

$\frac{1}{2}$

answer is C.

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Detailed Solution

$n (S) = 2^{10}$

E=getting as many heads in the 5 tosses as in the last 5 tosses

$n (E) = {}^{5}{C_{0}} {}^{5}{C_{0}} + {}^{5}{C_{1}} {}^{5}{C_{1}} + . . . . + {}^{5}{C_{5}} {}^{5}{C_{5}} = {}^{10}{C_{5}}$

$\Rightarrow P (E) = \frac{{}^{10}{C_{5}}}{2^{10}}$

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