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 a tail then a card from a well-shuffled pack of nine cards numbered $1,2,3,.......................,9$ is randomly picked and the number is either 7 or 8 is?
A. $\dfrac{{13}}{{36}}$
B. $\dfrac{{19}}{{36}}$
C. $\dfrac{{19}}{{72}}$
D. $\dfrac{{15}}{{72}}$

Answer 1

First of all, find the probability of getting a head and a tail. Then use the concepts of mutually exclusive events and the set of mutually independent events to reach the solution of the given problem.

Complete step by step answer:
Let $P\left( H \right)$ be the probability of getting a head and $P\left( T \right)$is getting a tail.
We know that for an unbiased coin probability of getting a head and probability of getting a tail are equal to $\dfrac{1}{2}$.
Therefore, $P\left( H \right) = P\left( T \right) = \dfrac{1}{2}$.
Now, let ${E_1}$ be the event of getting a sum of 7 or 8 when a pair of dice is thrown.
So, the total number of outcomes for event ${E_1} = 6 \times 6 = 36$.
We know that, when two dices are rolled the possible outcomes are

$$\left( {1,1} \right),\left( {1,2} \right),\left( {1,3} \right),\left( {1,4} \right),\left( {1,5} \right),\left( {1,6} \right) \\\ \left( {2,1} \right),\left( {2,2} \right),\left( {3,3} \right),\left( {4,4} \right),\left( {5,5} \right),\left( {6,6} \right) \\\ \left( {3,1} \right),\left( {3,2} \right),\left( {3,3} \right),\left( {3,4} \right),\left( {3,5} \right),\left( {3,6} \right) \\\ \left( {4,1} \right),\left( {4,2} \right),\left( {4,3} \right),\left( {4,4} \right),\left( {4,5} \right),\left( {4,6} \right) \\\ \left( {5,1} \right),\left( {5,2} \right),\left( {5,3} \right),\left( {5,4} \right),\left( {5,5} \right),\left( {5,6} \right) \\\ \left( {6,1} \right),\left( {6,2} \right),\left( {6,3} \right),\left( {6,4} \right),\left( {6,5} \right),\left( {6,6} \right) \\\$$

Out of these outcomes, the possibilities of getting a sum of 7 or 8 on the two dice are $\left( {1,6} \right),\left( {2,5} \right),\left( {2,6} \right),\left( {3,4} \right),\left( {3,5} \right),\left( {4,3} \right),\left( {4,4} \right),\left( {5,2} \right),\left( {5,3} \right),\left( {6,1} \right),\left( {6,2} \right)$
So, the number of possible outcomes for event ${E_1} = 11$
We know that the probability of an event $E$ is given by $P\left( E \right) = \dfrac{{{\text{Number of favorable outcomes}}}}{{{\text{Total number of outcomes}}}}$
Thus, $P\left( {{E_1}} \right) = \dfrac{{11}}{{36}}$
Let ${E_2}$ be the event of getting 7 or 8 when a card is picked from $1,2,3,............,9$
So, the total number of outcomes for ${E_2} = 9$
The possibilities of getting 7 or 8 from cards are 7 and 8 numbered cards only.
So, the number of possible outcomes for event ${E_2} = 2$
Thus, $P\left( {{E_2}} \right) = \dfrac{2}{9}$
Now, the probability of that the noted number is either 7 or 8 $= P\left( {\left( {H \cap {E_1}} \right){\text{ or }}\left( {T \cap {E_2}} \right)} \right)$

= P\left( {H \cap {E_1}} \right) + P\left( {T \cap {E_2}} \right){\text{ }}\left[ {\because \left\\{ {H,{E_1}} \right\\}{\text{ and }}\left\\{ {T,{E_2}} \right\\}{\text{ are mutually exclusive events}}} \right] \\\ = P\left( H \right).P\left( {{E_1}} \right) + P\left( T \right).P\left( {{E_2}} \right){\text{ }}\left[ {\because \left\\{ {H,{E_1}} \right\\}{\text{ and }}\left\\{ {T,{E_2}} \right\\}{\text{ are sets of independent events}}} \right] \\\ = \dfrac{1}{2} \times \dfrac{{11}}{{36}} + \dfrac{1}{2} \times \dfrac{2}{9} \\\ = \dfrac{{11}}{{72}} + \dfrac{2}{{18}} \\\ = \dfrac{{11 + 8}}{{72}} = \dfrac{{19}}{{72}} \\\

Hence, the probability that the noted number is either 7 or 8 is $\dfrac{{19}}{{72}}$.

So, the correct answer is “Option C”.

Note: Unbiased coin means that the probability of the heads is the same as the probability of the tails, each being $\dfrac{1}{2}$ (equal probability of selection). Two events are said to be mutually exclusive or disjoint if they cannot both occur at the same time. The set of events is mutually independent if each event is independent of each intersection of the other events.