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(B) Let $H$ be the event of getting a head and $T$ be the event of getting a tail. $P(H) = P(T) = \frac{1}{2}$.Case $1$: If $H$ occurs,two dice are ro

Vedclass · Accepted Answer

$0.244$

Vedclass · Answer

(B) Let $H$ be the event of getting a head and $T$ be the event of getting a tail. $P(H) = P(T) = \frac{1}{2}$.Case $1$: If $H$ occurs,two dice are rolled. The sum can be $7$ or $8$.$P(7|H) = \frac{6}{36} = \frac{1}{6}$ and $P(8|H) = \frac{5}{36}$.Case $2$: If $T$ occurs,a card is picked from $11$ cards numbered $2$ to $12$.$P(7|T) = \frac{1}{11}$ and $P(8|T) = \frac{1}{11}$.Total probability of getting $7$ is $P(7) = P(H)P(7|H) + P(T)P(7|T) = \frac{1}{2} 	imes \frac{1}{6} + \frac{1}{2} 	imes \frac{1}{11} = \frac{1}{2} \left( \frac{1}{6} + \frac{1}{11} ight) = \frac{1}{2} \left( \frac{17}{66} ight) = \frac{17}{132}$.Total probability of getting $8$ is $P(8) = P(H)P(8|H) + P(T)P(8|T) = \frac{1}{2} 	imes \frac{5}{36} + \frac{1}{2} 	imes \frac{1}{11} = \frac{1}{2} \left( \frac{5}{36} + \frac{1}{11} ight) = \frac{1}{2} \left( \frac{55 + 36}{396} ight) = \frac{91}{792}$.Required probability $P = P(7) + P(8) = \frac{17}{132} + \frac{91}{792} = \frac{102 + 91}{792} = \frac{193}{792} \approx 0.24368 \approx 0.244$.

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