Three six faced fair dice are thrown together. The probability that the sum of the numbers appearing on the dice is k(3 ≤ k ≤ 8) , is
Answer & explanation
Correct answer: option 1
The total number of cases = 6 × 6 × 6 = 216
The number of favourable ways
= Coefficient of $x^k $ in $(x + x^2 + ... + x^6)^3$
= Coefficient of $x^{k-3}$ in $ (1-x^6)^3 (1-x)^{-3}$
= Coefficient of $x^{k-3}$ in $ (1-x)^{-3} $ {0 ≤ k - 3 ≤ 5}
= Coefficient of $x^{k-3}$ in $(1+ {^3C}_1 x + {^4C}_2 x^2 + {^5C}_3 x^3 + ...) = {^{k-1}C}_2 =\frac{(k-1)(k-2)}{2}$
Thus the probability of the required event is $\frac{(k-1)(k-2)}{432}$.