Question

A fair coin is tossed. If the result is head, a pair of fair dice is rolled and the number obtained by adding the numbers on the two faces is noted. If the result is tail,a card from a well shuffled pack of $11$ cards is noted. The probability that the noted number is $7$ or $8,$ is :

• A. $193/794$
• B. $195/792$
• C. $191/792$
• D. None of these

Answer 1

The correct option is A $193/794$

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. of 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

$P\left(E_1\right)=\frac{1}{2}\times \frac{1}{6}+\frac{1}{2}\times \frac{1}{11}=\frac{17}{132}$

$P\left(E_2\right)=\frac{1}{2}\times \frac{5}{36}+\frac{1}{2}\times \frac{1}{11}$$=\frac{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