Question

An unbiased coin is tossed. If the outcome is a head then a pair of unbiased dice is rolled and the sum of the numbers obtained on them is noted. If the toss of the coin results in tail then a card from a well-shuffled pack of nine cards numbered 1, 2, 3, …., 9 is randomly picked and the number on the card is noted. The probability that the noted number is either 7 or 8 is:
1. 13/36
2. 15/72
3. 19/72
4. 19/36

Answer 1

Correct Answer - Option 3 : 19/72

From question, we need to find the probability of the number either 7 or 8 in dice and card when the coin is head and tail respectively.

Probability of getting head:

\(\Rightarrow P\left( H \right) = \frac{1}{2}\)

Probability of getting tail:

\(\Rightarrow P\left( T \right) = \frac{1}{2}\)

Probability of getting sum of 7 or 8 when the dice is rolled:

\(\Rightarrow P\left( {{E_1}} \right) = \frac{6}{{36}} + \frac{5}{{36}} = \frac{{11}}{{36}}\)

Probability of getting 7 or 8 in the shuffled card is:

\(\Rightarrow P\left( {{E_2}} \right) = \frac{1}{9} + \frac{1}{9} = \frac{2}{9}\)

Now, from the question,

⇒ P(H ∩ E₁) + P(T∩E₂)

∵ {H, E₁} and {T, E₂} are sets of independent events.

⇒ [P(H) × P(E₁)] + [P(T) × P(E₂)]

On substituting the values,

\(\Rightarrow \left[ {\frac{1}{2} \times \frac{{11}}{{36}}} \right] + \left[ {\frac{1}{2} \times \frac{2}{9}} \right]\)

\(\Rightarrow \frac{{11}}{{72}} + \frac{2}{{18}}\)

\(\Rightarrow \frac{{11 + 2\left( 4 \right)}}{{72}}\)

\(\Rightarrow \frac{{11 + 8}}{{72}}\)

\(\therefore P\left( {H \cap {E_1}} \right) + P\left( {T \cap {E_2}} \right) = \frac{{19}}{{72}}\)