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

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(C) The random variable $X$ represents the number of sixes in $2$ tosses of a fair die. The possible values for $X$ are $0, 1, 2$.The probability of g

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$X = x$$0$$1$$2$$P(X = x)$$\frac{25}{36}$$\frac{5}{18}$$\frac{1}{36}$

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(C) The random variable $X$ represents the number of sixes in $2$ tosses of a fair die. The possible values for $X$ are $0, 1, 2$.The probability of getting a six in a single toss is $p = \frac{1}{6}$,and the probability of not getting a six is $q = 1 - \frac{1}{6} = \frac{5}{6}$.For $X = 0$ (no sixes): $P(X = 0) = q 	imes q = \frac{5}{6} 	imes \frac{5}{6} = \frac{25}{36}$.For $X = 1$ (one six): $P(X = 1) = (p 	imes q) + (q 	imes p) = (\frac{1}{6} 	imes \frac{5}{6}) + (\frac{5}{6} 	imes \frac{1}{6}) = \frac{5}{36} + \frac{5}{36} = \frac{10}{36} = \frac{5}{18}$.For $X = 2$ (two sixes): $P(X = 2) = p 	imes p = \frac{1}{6} 	imes \frac{1}{6} = \frac{1}{36}$.The probability distribution is:$X = x$$0$$1$$2$$P(X = x)$$\frac{25}{36}$$\frac{5}{18}$$\frac{1}{36}$

$X=x$	$1$	$2$	$3$	$4$
$P(X=x)$	$\frac{2}{20}$	$\frac{4}{20}$	$\frac{6}{20}$	$\frac{8}{20}$

$X$	$1$	$2$	$3$	$4$	$5$	$6$
$F(X=x)$	$0.2$	$0.37$	$0.48$	$0.62$	$0.85$	$1$

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

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Then $P[X=4] + P[X=5] = $

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$X = x$	$0$	$1$	$2$
$P(X = x)$	$\frac{25}{36}$	$\frac{1}{36}$	$\frac{5}{18}$