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

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Question

(A) Let $p$ be the probability of getting a six in a single toss,so $p = \frac{1}{6}$.Let $q$ be the probability of not getting a six,so $q = 1 - \fra

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $p$ be the probability of getting a six in a single toss,so $p = \frac{1}{6}$.Let $q$ be the probability of not getting a six,so $q = 1 - \frac{1}{6} = \frac{5}{6}$.$X$ denotes the number of sixes in two tosses. $X$ can take values $0, 1, 2$.$P(X = 0) = P(	ext{no six}) = q 	imes q = \frac{5}{6} 	imes \frac{5}{6} = \frac{25}{36}$.$P(X = 1) = P(	ext{one six}) = pq + qp = \frac{1}{6} 	imes \frac{5}{6} + \frac{5}{6} 	imes \frac{1}{6} = \frac{5}{36} + \frac{5}{36} = \frac{10}{36} = \frac{5}{18}$.$P(X = 2) = P(	ext{two sixes}) = p 	imes p = \frac{1}{6} 	imes \frac{1}{6} = \frac{1}{36}$.Thus,the probability distribution is:$X = x_i$$0$$1$$2$$P(x_i)$$\frac{25}{36}$$\frac{5}{18}$$\frac{1}{36}$

$X = x_i$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
$P_i$	$\frac{1}{6}$	$\frac{1}{12}$	$\frac{1}{12}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$

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

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