An unbiased coin is tossed. If the result is a head, a pair of unbiase

1.	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?
Answer» 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

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?

Answer»

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