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 number on the two faces is noted. If the result is a tail. a card from a well shufled pack of eleven cards numbered $2,3,4,...,12$ is picked and the number on the card is noted. The probability that the noted number is either $7$ or $8,$ is $\frac{k}{729}$, then the last digit of $k$ is

Answer 1

The possible outcomes on the pair of dice when the sum of the numbers on them is $7$ are $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1),$ the probability for which is $\frac{6}{36}$.
The possible outcomes on the dice when the sum is $8$ are $(2,5),(3,5),(4,4,),(5,3),(6,2)$ the probability for which is $\frac{5}{36}.$
$\therefore$ The probability that on the coin head comes and on the dice the sum is $7$ or $8$ $=\frac{1}{2}(\frac{6}{36}+\frac{5}{36})=\frac{11}{72}$
The probability that from the pack of $11$ cards when one card is chosen then number on it is $7$ or $8$ $=\frac{2}{11}$
$\therefore$ The probability that from tail comes on the coin and then on the chosen card number appearing is $7$ or $8$
$=\frac{1}{2}\times \frac{2}{11}=\frac{1}{11}.$
Thus, the required probability
$=\frac{11}{72}+\frac{1}{11}=\frac{121+72}{11\times 72}=\frac{193}{792}.$

The last digit of $k$ is $3$.