Explanation

(b)
Let $p = P$(getting a head)$= \frac{1}{2},\ q = P$(getting no head)$= \frac{1}{2}$
By using binomial distribution,
Required probability $P$(six heads)
$= {\ ^{10}C}_{6}\left( \frac{1}{2} \right)^{6}\left( \frac{1}{2} \right)^{4} = \frac{10!}{6!4!} \times \frac{1}{2^{10}}$
$= \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2} \times \frac{1}{2^{10}} = \frac{105}{512}$