# 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, . . . . . . . . . . . . . . . . 12 are picked and the number on the card is noted. What is the probability that the noted number is either 7 or 8?

8 years ago
Hello Student,
E1 ≡ number noted is 7, E2 ≡ number notes is 8,
H ≡ getting head on coin, T ≡ getting tail on coin.
Then by total probability theorem,
P (E1) = P (H) P (E1| H) + P (T) P (E1| T)
And P (E2) = P (H) P (E2| H) + P (T) P (E2| T)
Where P (H) = 1/2 ; P (T) = 1/2
P (E1| H) = prob. of getting a sum of 7 on two dice. Here favorable cases are
{(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)}
∴ P (E1| H) = 6/36 = 1/6
Also P (E1| T) = prob. if getting ‘7’ numbered card out of 11 cards = 1/11.
P (E2| H) = Prob. of getting a sum of 8 on two dice. Here favorable cases are
{(2, 6), (6, 2), (4, 4), (5, 3), (3, 5)}
∴ P (E2| H) = 5/36
P (E2| T) = prob. of getting ‘8’ numbered card out of 11 cards = 1/11
∴ P (E1) = 1/2 x 1/6 + 1/2 x 1/11 = 1/12 + 1/22 = 11 + 6/132 = 17/132
P (E2) = 1/2 x 5/36 + 1/2 x 1/11 = 1/2 [55 + 36/396] = 91/792
Now E1 and E2 are mutually exclusive events therefore
P (E1 or E2) = P (E1) + P (E2) = 17/132 + 91/792
= 102 + 91/792 = 193/792 = 0.2436

Thanks