उत्तर (12)

समाधान

$a=P(X=3)=\dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{1}{6}=\dfrac{25}{216}$

$b=P(X \geq 3)=P(X=4,5,6,7….)=\dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{1}{6}+\left(\dfrac{5}{6}\right)^{3} \cdot \dfrac{1}{6}+\left(\dfrac{5}{6}\right)^{4} \cdot \dfrac{1}{6}+\ldots \ldots$

$=\dfrac{\dfrac{25}{216}}{1-\dfrac{5}{6}}=\dfrac{25}{216} \times \dfrac{6}{1}=\dfrac{25}{36}$

$c=P(X \geq 6 \mid X>3)= \dfrac{ P(X\geq6\cap X >3)}{P(X>3)}$

$=\dfrac{ P(X \geq 6}{P(X>3)}$

$P(X \geq 6)=\left(\dfrac{5}{6}\right)^{5} \cdot \dfrac{1}{6}+\left(\dfrac{5}{6}\right)^{6} \cdot \dfrac{1}{6}+\ldots \ldots$.

$=\dfrac{\left(\dfrac{5}{6}\right)^{5} \cdot \dfrac{1}{6}}{1-\dfrac{5}{6}}=\left(\dfrac{5}{6}\right)^{5}$

$c=\dfrac{\left(\dfrac{5}{6}\right)^{5}}{\left(\dfrac{5}{6}\right)^{3}}=\dfrac{25}{36}$

$\dfrac{b+c}{a}=\dfrac{\left(\dfrac{5}{6}\right)^{2}+\left(\dfrac{5}{6}\right)^{2}}{\left(\dfrac{5}{6}\right)^{2} \cdot \dfrac{1}{6}}=12$