Explanation

(d)
The probability of getting a number either 2 or 3 in one toss is $\frac{1}{2}$
Condition for getting the sum 12 in five tossed is (2, 2, 2, 3, 3).
$\therefore$ Required probability $= {\ ^{5}C}_{3}\left( \frac{1}{2} \right)^{3}\left( \frac{1}{2} \right)^{2}$
$= \frac{5 \times 4}{2 \times 1}\left( \frac{1}{2} \right)^{5}$
$= 10 \bullet \frac{1}{2^{5}} = \frac{5}{16}$