Question

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 $2,3,4,....,12$ is picked and the number on the card is noted. The probability that the noted number is either $7$ or $8$, is

1. $0.24$
2. $0.244$
3. $0.024$
4. None of these

Answer 1

Hint : Here, to solve the problem we will use the concept of total probability. The formula we will use will be,
$P\left( E \right) = P\left( H \right).P\left( {\dfrac{E}{H}} \right) + P\left( T \right).P\left( {\dfrac{E}{T}} \right)$
So, here we have to find the probability for two outcomes, one is for $7$ and another for $8$. So, by applying the formula of total probability for both, then we have to see whether the events are mutually exclusive or not. Mutually exclusive events are events that cannot happen at the same time. They don’t have anything in common.
If mutually exclusive, then we can just simply add the probability of the two events and easily find the value of the total probability of the two events just by simply their probability.

Complete step-by-step answer :
Let ${E_1}$ be the event that the number noted at the end is $7$.
And, ${E_2}$ be the event that the number noted at the end is $8$.
Let $H$ be the event that the head occurs on a coin.
And, $T$ be the event that the tail occurs on the coin.
Therefore, using the Bayes theorem and the law of total probability, we know,
$P\left( {{E_1}} \right) = P\left( H \right).P\left( {\dfrac{{{E_1}}}{H}} \right) + P\left( T \right).P\left( {\dfrac{{{E_1}}}{T}} \right)$
And, $P\left( {{E_2}} \right) = P\left( H \right).P\left( {\dfrac{{{E_2}}}{H}} \right) + P\left( T \right).P\left( {\dfrac{{{E_2}}}{T}} \right)$
Now, we know, in an unbiased coin, the probability of head and tail are equal.
Therefore, $P\left( H \right) = P\left( T \right) = \dfrac{1}{2}$
Now, if head occurs on the coin first, then, a pair of unbiased dice is rolled.
So, the total number outcomes on the dice will be $= 36$
Now, number of outcomes such that the sum of numbers on the two dice is $7$are,
\left\\{ {\left( {1,6} \right),\left( {6,1} \right),\left( {2,5} \right),\left( {5,2} \right),\left( {3,4} \right),\left( {4,3} \right)} \right\\}
Therefore, the probability of getting a sum of $7$ on the two dice, $P\left( {\dfrac{{{E_1}}}{H}} \right) = \dfrac{6}{{36}} = \dfrac{1}{6}$
Now, if a tail occurs on the coin first, then, a card from a well shuffled pack of eleven cards numbered $2,3,4,....,12$ is picked.
So, total number of cards $= 11$
Number of favourable outcomes of getting the number $7$ on the card $= 1$
Therefore, the probability of getting the number $7$ on the card, $P\left( {\dfrac{{{E_1}}}{T}} \right) = \dfrac{1}{{11}}$.
Also, number of outcomes such that the sum of numbers on the two dice is $8$are,
\left\\{ {\left( {2,6} \right),\left( {6,2} \right),\left( {3,5} \right),\left( {5,3} \right),\left( {4,4} \right)} \right\\}
Therefore, the probability of getting a sum of $8$ on the two dice, $P\left( {\dfrac{{{E_2}}}{H}} \right) = \dfrac{5}{{36}}$
Number of favourable outcomes of getting the number $8$ on the card $= 1$
Therefore, the probability of getting the number $8$ on the card, $P\left( {\dfrac{{{E_2}}}{T}} \right) = \dfrac{1}{{11}}$
Now, substituting these values to find the values of $P\left( {{E_1}} \right)$ and $P\left( {{E_2}} \right)$
$\therefore P\left( {{E_1}} \right) = \left( {\dfrac{1}{2}} \right)\left( {\dfrac{1}{6}} \right) + \left( {\dfrac{1}{2}} \right)\left( {\dfrac{1}{{11}}} \right)$
$\Rightarrow P\left( {{E_1}} \right) = \dfrac{1}{{12}} + \dfrac{1}{{22}}$
Taking LCM, we get,
$\Rightarrow P\left( {{E_1}} \right) = \dfrac{{11 + 6}}{{132}}$
$\Rightarrow P\left( {{E_1}} \right) = \dfrac{{17}}{{132}}$
And, $P\left( {{E_2}} \right) = \left( {\dfrac{1}{2}} \right)\left( {\dfrac{5}{{36}}} \right) + \left( {\dfrac{1}{2}} \right)\left( {\dfrac{1}{{11}}} \right)$
$\Rightarrow P\left( {{E_2}} \right) = \dfrac{5}{{72}} + \dfrac{1}{{22}}$
Taking LCM, we get,
$\Rightarrow P\left( {{E_2}} \right) = \dfrac{{55 + 36}}{{792}}$
$\Rightarrow P\left( {{E_2}} \right) = \dfrac{{91}}{{792}}$
We can easily say that, the two events are mutually exclusive, hence,
Probability of getting $7$ or $8$ $= P\left( {{E_1}{\text{ or }}{E_2}} \right) = P\left( {{E_1}} \right) + P\left( {{E_2}} \right)$
$= \dfrac{{17}}{{132}} + \dfrac{{91}}{{792}}$
Taking LCM, we get,
$= \dfrac{{102 + 91}}{{792}}$
$= \dfrac{{193}}{{792}}$
$= 0.24$
Therefore, the correct option is 1.
So, the correct answer is “Option 1”.

Note : In this problem, the two events were mutually exclusive, so it was easy for us to find the total probability. But if the two events were mutually inclusive then a term of outcomes that suits both the events, i.e., the probability of the intersection of the two events have to be taken into account and subtracted from the sum of the probability of the two events in order to get the answer.