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

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(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}$

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

$X$	$4k$	$\frac{30}{7}k$	$\frac{32}{7}k$	$\frac{34}{7}k$	$\frac{36}{7}k$	$\frac{38}{7}k$	$\frac{40}{7}k$	$6k$
$P(X)$	$\frac{2}{15}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{1}{15}$

$X=x$	$-3$	$6$	$9$
$P(X=x)$	$\frac{1}{6}$	$\frac{1}{2}$	$\frac{1}{3}$

$X=x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X=x)$	$0$	$k$	$2k$	$2k$	$3k$	$k^2$	$2k^2$	$7k^2+k$

$A$ 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

Similar Questions

$A$ fair six-faced die is rolled $12$ times. The probability that each face turns up exactly twice is equal to:

If $X$ is a random variable with the following probability distribution:
$X=x$ $-3$ $6$ $9$
$P(X=x)$ $\frac{1}{6}$ $\frac{1}{2}$ $\frac{1}{3}$

Then the variance of $X$ is:

If a random variable $X$ has the following probability distribution of $X$:

$X=x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$

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Then $P(X \geqslant 6) = $

$A$ fair die is tossed repeatedly until a six is obtained. Let $X$ denote the number of tosses required and let $a=P(X=3)$,$b=P(X \geq 3)$ and $c=P(X \geq 6 \mid X>3)$. Then $\frac{b+c}{a}$ is equal to

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