A fair die is tossed twice in succession. If | vedclass

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(C) fair die is tossed twice in succession. The total number of outcomes is $6 	imes 6 = 36$.Let $X$ be the number of fours in $2$ tosses.The possibl

Vedclass · Accepted Answer

$X = x_i$$0$$1$$2$$P_i$$\frac{25}{36}$$\frac{5}{18}$$\frac{1}{36}$

Vedclass · Answer

(C) fair die is tossed twice in succession. The total number of outcomes is $6 	imes 6 = 36$.Let $X$ be the number of fours in $2$ tosses.The possible values for $X$ are $0, 1, 2$.$1$. For $X = 0$: The outcomes are all pairs where neither die shows a $4$. There are $5 	imes 5 = 25$ such outcomes. Thus,$P(X = 0) = \frac{25}{36}$.$2$. For $X = 1$: The outcomes are $(4, 	ext{not } 4)$ or $(	ext{not } 4, 4)$. There are $1 	imes 5 + 5 	imes 1 = 10$ such outcomes. Thus,$P(X = 1) = \frac{10}{36} = \frac{5}{18}$.$3$. For $X = 2$: The only outcome is $(4, 4)$. There is $1$ such outcome. Thus,$P(X = 2) = \frac{1}{36}$.The probability distribution is:$X = x_i$$0$$1$$2$$P_i$$\frac{25}{36}$$\frac{5}{18}$$\frac{1}{36}$

No. of Pizzas $(x_i)$	$5$	$6$	$7$	$8$	$9$	$10$
Probability $(P_i)$	$0.07$	$0.2$	$0.3$	$0.3$	$0.07$	$0.06$

$A$ fair die is tossed twice in succession. If $X$ denotes the number of fours in $2$ tosses,then the probability distribution of $X$ is given by

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$X = x_i$	$0$	$1$	$2$
$P_i$	$\frac{1}{36}$	$\frac{25}{36}$	$\frac{5}{18}$