Explanation

(c)
(a) $P$ (no boy in family of 4) $= P$ ( all girls in it)$= \left( \frac{1}{2} \right)^{4} = \frac{1}{16}$
Hence, the probability of having at least one boy
$= 1 - \frac{1}{16} = \frac{15}{16}$
(b)$P$ (first card is an ace)$= \frac{1}{13}$
and $P$ (second card is an ace) $= \frac{1}{13}$
Therefore, $P$ (both cards are aces) $= \frac{1}{13} \times \frac{1}{13} = \frac{1}{169}$
(c) Let guessing correctly one answer as a success. Then, we have
$p = \frac{1}{2},q = \frac{1}{2},n = 10$
$\therefore\ P(8) + P(9) + P(10)$
$= {\ ^{10}C}_{8}\ \left( \frac{1}{2} \right)^{8}\left( \frac{1}{2} \right)^{2} + {\ ^{10}C}_{9}\left( \frac{1}{2} \right)^{9}\left( \frac{1}{2} \right) + {\ ^{10}C}_{10}\left( \frac{1}{2} \right)^{10}$
$= \frac{45 + 10 + 1}{1024} = \frac{7}{128}$
(d) we have, $n = 3,\ p = \frac{1}{2},\ q = \frac{1}{2}$
Where obtaining a head has been reckoned a success.
Now, $P(2) = {\ ^{3}C}_{2}\left( \frac{1}{2} \right)^{2}\left( \frac{1}{2} \right)^{1} = \frac{3}{8}$
Hence, it is clear that option (c) is not correct.