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(C) The total number of outcomes when three dice are thrown is $n(S) = 6 	imes 6 	imes 6 = 216$. Let $x, y, z$ be the outcomes on the three dice,whe

Vedclass · Accepted Answer

$\frac{5}{72}$

Vedclass · Answer

(C) The total number of outcomes when three dice are thrown is $n(S) = 6 	imes 6 	imes 6 = 216$. Let $x, y, z$ be the outcomes on the three dice,where $1 \le x, y, z \le 6$. We need to find the number of integer solutions to $x + y + z = 7$. This is equivalent to finding the coefficient of $x^7$ in the expansion of $(x + x^2 + x^3 + x^4 + x^5 + x^6)^3$. $= x^3(1 + x + x^2 + x^3 + x^4 + x^5)^3 = x^3 \left( \frac{1 - x^6}{1 - x} ight)^3$. We need the coefficient of $x^7$ in $x^3(1 - x^6)^3(1 - x)^{-3}$,which is the coefficient of $x^4$ in $(1 - 3x^6 + 3x^{12} - x^{18})(1 - x)^{-3}$. Using the binomial expansion $(1 - x)^{-3} = \sum_{r=0}^{\infty} \binom{r+3-1}{3-1} x^r = \sum_{r=0}^{\infty} \binom{r+2}{2} x^r$,the coefficient of $x^4$ is $\binom{4+2}{2} = \binom{6}{2} = \frac{6 	imes 5}{2} = 15$. Therefore,the probability is $P(E) = \frac{n(E)}{n(S)} = \frac{15}{216} = \frac{5}{72}$.

If three dice are thrown simultaneously,then the probability of getting a score of $7$ is

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