Four fair dice D_1, D_2, D_3 and D_4,each hav | vedclass

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(A) Let $X$ be the number shown by $D_4$. The total number of outcomes for the four dice is $6^4 = 1296$.Alternatively,consider the probability for a

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$\frac{91}{216}$

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(A) Let $X$ be the number shown by $D_4$. The total number of outcomes for the four dice is $6^4 = 1296$.Alternatively,consider the probability for a fixed value of $D_4$. For any specific number $k \in \{1, 2, 3, 4, 5, 6\}$ shown by $D_4$,the probability that $k$ does $NOT$ appear on any of $D_1, D_2, D_3$ is $(\frac{5}{6})^3 = \frac{125}{216}$.Therefore,the probability that $k$ appears on at least one of $D_1, D_2, D_3$ is $1 - \frac{125}{216} = \frac{216 - 125}{216} = \frac{91}{216}$.Since this probability is independent of the value shown by $D_4$,the overall probability is $\frac{91}{216}$.

$Face$	$1$	$2$	$3$	$4$	$5$	$6$
$P(F)$	$0.1$	$0.24$	$0.19$	$0.18$	$0.15$	$0.14$

$X$	$0$	$1$	$2$	$3$	$4$
$P(X)$	$k$	$2k$	$4k$	$6k$	$8k$

Four fair dice $D_1, D_2, D_3$ and $D_4$,each having six faces numbered $1, 2, 3, 4, 5$ and $6$,are rolled simultaneously. The probability that $D_4$ shows a number appearing on at least one of $D_1, D_2$ and $D_3$ is

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