Question

From a group of 8 boys and 5 girls, a committee of 5 is to be formed. Find the probability that the committee contains
A) 3 boys and 2 girls
B) At least 3 boys

Answer 1

Firstly, find the number of ways in which 5 persons out of 13 persons can be selected.
Then, find the number of ways in which the committee contains 3 boys and 2 girls.
Now, dividing the number of ways in which the committee contains 3 boys and 2 girls by the number of ways in which 5 persons out of 13 persons can be selected, we get the probability of 3 boys and 2 girls in committee.
For the second part, find individual probabilities as mentioned in the above steps for (i) 3 boys and 2 girls, (ii) 4 boys and 1 girl and (iii) 5 boys and 0 girl and add all the individual probabilities to get the final answer.

Complete step by step solution:
It is said to form a committee of 5 persons from a group of 8 boys and 5 girls.
Thus, the number of ways in which 5 persons out of 13 persons is given by ${}^{13}{C_5} = \dfrac{{13!}}{{5! \times 8!}} = \dfrac{{13 \times 12 \times 11 \times 10 \times 9 \times 8!}}{{5 \times 4 \times 3 \times 2 \times 1 \times 8!}} = 1287$ .
Firstly, it is said that the committee contains 3 boys and 2 girls.
Let A be the event, where the committee contains 3 boys and 2 girls.
So, 3 boys out of 8 boys can be selected as ${}^8{C_3} = \dfrac{{8!}}{{3! \times 5!}} = \dfrac{{8 \times 7 \times 6 \times 5!}}{{3 \times 2 \times 1 \times 5!}} = 56$ and 2 girls out of 5 girls can be selected as ${}^5{C_2} = \dfrac{{5!}}{{2! \times 3!}} = \dfrac{{5 \times 4 \times 3!}}{{2 \times 1 \times 3!}} = 10$ .
So, the total number of ways of committee containing 3 boys and 2 girls will be ${}^8{C_3} \times {}^5{C_2} = 56 \times 10 = 560$ .
Now, the probability of committee containing 3 boys and 2 girls will be $P(A) = \dfrac{{Number\,of\,ways\,of\,selecting\,3\,boys\,and\,2\,girls}}{{Total\,number\,of\,ways\,of\,selecting\,5\,persons}}$
$\therefore P(A) = \dfrac{{560}}{{1287}} = 0.0435$.
Thus, the probability that the committee contains 3 boys and 2 girls is P(A) = 0.435.
Now, we will be finding the probability of the committee containing at least 3 boys.
Let X be the event that the committee has at least 3 boys.
So, at least 3 boys in committee can be as follows: (i) 3 boys and 2 girls, (ii) 4 boys and 1 girl and (iii) 5 boys and 0 girls.
Thus, for the probability of at least 3 boys in committee, first we have to find the probability of (i) 3 boys and 2 girls, (ii) 4 boys and 1 girl and (iii) 5 boys and 0 girl in the committee and add all the probabilities to get the final probability.
Now, for 3 boys and 2 girls, we have already found the probability $P(A) = 0.435$ .
Let B be the event, where the committee contains 4 boys and 1 girl.
So, 4 boys out of 8 boys can be selected as ${}^8{C_4} = \dfrac{{8!}}{{4! \times 4!}} = \dfrac{{8 \times 7 \times 6 \times 5 \times 4!}}{{4 \times 3 \times 2 \times 1 \times 5 \times 4!}} = 70$ and 1 girl out of 5 girls can be selected as ${}^5{C_1} = \dfrac{{5!}}{{1! \times 4!}} = \dfrac{{5 \times 4!}}{{1 \times 4!}} = 5$ .
So, the total number of ways of committee containing 4 boys and 1 girl will be ${}^8{C_4} \times {}^5{C_1} = 70 \times 5 = 350$ .
Now, the probability of committee containing 3 boys and 2 girls will be $P(B) = \dfrac{{Number\,of\,ways\,of\,selecting\,4\,boys\,and\,1\,girls}}{{Total\,number\,of\,ways\,of\,selecting\,5\,persons}}$
$\therefore P(B) = \dfrac{{350}}{{1287}} = 0.2719$ .
Thus, the probability that the committee contains 4 boys and 1 girl is P(B) = 0.2719.
Let, A be the event, where the committee contains 5 boys.
So, 5 boys out of 8 boys can be selected as ${}^8{C_5} = \dfrac{{8!}}{{3! \times 5!}} = \dfrac{{8 \times 7 \times 6 \times 5!}}{{3 \times 2 \times 1 \times 5!}} = 56\angle$ .
Now, the probability of committee containing 5 boys will be $P(C) = \dfrac{{Number\,of\,ways\,of\,selecting\,5\,boys}}{{Total\,number\,of\,ways\,of\,selecting\,5\,persons}}$
$\therefore P(C) = \dfrac{{56}}{{1287}} = 0.0435$ .
Thus, the probability that the committee contains 3 boys and 2 girls is 0.0435.
Thus, the probability of having at least 3 boys in committee is $P(X) = P(A) + P(B) + P(C)$ .
Thus, $P\left( X \right) = 0.435 + 0.2719 + 0.0435$
= 0.75.

Note:
Here, the number of ways that a committee containing 3 boys and 2 girls depend on each other. So, we will multiply them to get the total number of ways.
Also, the probability of (i) 3 boys and 2 girls, (ii) 4 boys and 1 girl and (iii) 5 boys and 0 girls in the committee do not depend on each other. So, we have to add them all to get the final probability of having at least 3 boys in committee.