Question

Three six-faced fair dice are thrown together. The probability that the sum of the numbers appearing on the dice is $k(3\le k\le 8)$ is

• A. $\frac{(k-1)(k-2)}{432}$
• B. $\frac{k(k-1)}{432}$
• C. ${}^{k-1}{C}_2\times \frac{1}{216}$
• D. $\frac{k^2}{432}$

Answer 1

The correct option is A $\frac{(k-1)(k-2)}{432}$

There are 3 dice having faces numbered 1 through 6. Thus, the favorable cases = coefficient of $x^k$ in the expression: ${(x+x^2+...+x^6)}^3=x^3(1+x+x^2+...+x^5)$
Or, the coefficient of $x^{k-3}$ in the expression: ${(1+x+x^2+...+x^5)}^3={(1-x^6)}^3{(1-x)}^{-3}$
Or, the coefficient of $x^{k-3}$ in the expression: ${(1-x)}^{-3}$ {since $(1-x^6)$ has powers of 6 and beyond while $0<k-3<5$}.
Or, the coefficient of $x^{k-3}$ in the expression: $1+C_1^{3}x+C_2^4{x}^2+C_3^5{x}^3+...$
= $C_{k-3}^{k-3+2}=C_{k-3}^{k-1}=C_2^{k-1}=\frac{(k-1)(k-2)}{2}$
Total cases = $6\ast 6\ast 6=216$
Hence, probability = $\frac{(k-1)(k-2)}{432}$
Alternate:
Probability of getting 3 as the sum = $\frac{1}{216}$ since 3 can be obtained in only 1 way: 1, 1, 1.
Substitute k = 3 in the options, only option (a) satisfies.
To be sure (since there is a 'none of these' option), we can take k = 4; which can be done in 3 ways: 1, 1, 2; or 1, 2, 1; or 2, 1, 1. So, probability = $\frac{3}{216}=\frac{1}{72}$