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

Vedclass · Accepted Answer

$\frac{91}{216}$

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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 $\frac{91}{216}$.

List-$I$	List-$II$
$A$. $P(E_2)$	$(i)$ $\frac{1}{2}$
$B$. $P(\frac{E_1}{E_2})$	$(ii)$ $\frac{5}{6}$
$C$. $P(\frac{\bar{E}_2}{E_1})$	$(iii)$ $\frac{1}{3}$
$D$. $P(\bar{E}_1 \cup \bar{E}_2)$	$(iv)$ $\frac{1}{6}$
	$(v)$ $\frac{2}{3}$

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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