Three six-faced fair dice are thrown together | vedclass

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(D) The total number of outcomes when three dice are thrown is $6 	imes 6 	imes 6 = 216$. To find the number of ways to get a sum $k$,we look for th

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(D) The total number of outcomes when three dice are thrown is $6 	imes 6 	imes 6 = 216$. To find the number of ways to get a sum $k$,we look for the coefficient of $x^k$ in the expansion of $(x + x^2 + x^3 + x^4 + x^5 + x^6)^3$. This is equal to the coefficient of $x^k$ in $x^3(1 + x + x^2 + x^3 + x^4 + x^5)^3 = x^3 \left( \frac{1 - x^6}{1 - x} ight)^3$. Since $3 \le k \le 8$,the term $(1 - x^6)$ does not affect the coefficient of $x^{k-3}$ in $(1 - x)^{-3}$. The coefficient of $x^{k-3}$ in $(1 - x)^{-3}$ is given by the formula $\binom{(k-3) + 3 - 1}{3 - 1} = \binom{k-1}{2}$. Calculating this,we get $\binom{k-1}{2} = \frac{(k-1)(k-2)}{2}$. Therefore,the probability is $\frac{\frac{(k-1)(k-2)}{2}}{216} = \frac{(k-1)(k-2)}{432}$.

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:

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