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(B) The random variable $X$ follows a geometric distribution with parameter $p = \frac{1}{6}$ and $q = \frac{5}{6}$.$a = P(X=3) = q^2 p = \left(\frac{

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$12$

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(B) The random variable $X$ follows a geometric distribution with parameter $p = \frac{1}{6}$ and $q = \frac{5}{6}$.$a = P(X=3) = q^2 p = \left(\frac{5}{6}ight)^2 	imes \frac{1}{6} = \frac{25}{216}$.$b = P(X \geq 3) = q^2 = \left(\frac{5}{6}ight)^2 = \frac{25}{36}$.For $c = P(X \geq 6 \mid X>3)$,by the memoryless property of the geometric distribution,$P(X \geq n+k \mid X>n) = P(X \geq k)$.Here,$n=3$ and $n+k=6$,so $k=3$.Thus,$c = P(X \geq 3) = q^2 = \left(\frac{5}{6}ight)^2 = \frac{25}{36}$.Finally,$\frac{b+c}{a} = \frac{\frac{25}{36} + \frac{25}{36}}{\frac{25}{216}} = \frac{\frac{50}{36}}{\frac{25}{216}} = \frac{50}{36} 	imes \frac{216}{25} = 2 	imes 6 = 12$.

$X$	$1$	$2$	$3$	$4$	$5$
$P(X=x)$	$K$	$2K$	$3K$	$2K$	$K$

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

$X$	$30$	$10$	$-10$
$P(X)$	$\frac{1}{5}$	$A$	$B$

$A$ 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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