An unbiased coin is tossed. If the result is a head a pair of unbiased dice is rolled and the number

Question

An unbiased coin is tossed. If the result is a head a pair of unbiased dice is rolled and the number obtained by adding the numbers on the two faces is noted. If the result is a tail, a card from a well shuffled pack of eleven cards numbered $2,3,4, \ldots \ldots ., 12$ is picked and the number is noted. The probability that the noted number is 7 or 8  is $\frac{m}{n}$ when expressed in the lowest form. Then m+n is?

Professor PC Thomas Classes · Answer

Probability of getting a head when a coin is tossed $=\frac{1}{2}$.\ Favourable cases of getting a sum of 7 is $\{(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)\}$\ Probability of getting 7 or 8 when two dice are rolled $=\frac{6}{36}+\frac{5}{36}=\frac{11}{36}$.\ Let $A$ be the event of getting 7 or 8 in this case.\ $	herefore \quad P(A)=\frac{1}{2} 	imes \frac{11}{36}$\ Probability of getting a tail when a coin is tossed $=\frac{1}{2}$\ Probability of getting 7 or 8 from the cards numbered 2 to $12=\frac{2}{11}$. Let $B$ the event of getting 7 or 8 in this case.\ $	herefore P(B)=\frac{1}{2} 	imes \frac{2}{11}$\ Required probability $=P(A)+P(B)$\ $=\frac{1}{2}\left(\frac{11}{36}+\frac{2}{11}ight)=\frac{193}{792}$.