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

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

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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 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}$

Outcome	Probability
$\omega_{1}$	$0.1$
$\omega_{2}$	$0.01$
$\omega_{3}$	$0.05$
$\omega_{4}$	$0.03$
$\omega_{5}$	$0.01$
$\omega_{6}$	$0.2$
$\omega_{7}$	$0.6$

$X = x$	$-1.5$	$-0.5$	$0.5$	$1.5$	$2.5$
$P(X = x)$	$0.05$	$0.2$	$0.15$	$0.25$	$0.35$

Outcome	$\omega_1$	$\omega_2$	$\omega_3$	$\omega_4$	$\omega_5$	$\omega_6$
$(e)$	$0.1$	$0.2$	$0.3$	$0.4$	$0.5$	$0.6$

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