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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,$\frac{b+c}{a} = \frac{\frac{25}{36} + \frac{25}{36}}{\frac{25}{216}} = \frac{\frac{50}{36}}{\frac{25}{216}} = \frac{50}{36} 	imes \frac{216}{25} = 2 	imes 6 = 12$.

$A$ 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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