Question

There are three boys and two girls. A committee of two is to be formed. Find the probability of the event that the committee contains at least one girl.

Answer 1

Hint: We use the principle of permutations and combinations to find the number of ways in which we can find the number of ways in which a committee of two can be formed from three boys and two girls. We will then use the concepts of probability to find the desired answer by calculating the desired number of outcomes (committee containing at least one girl). We will also make use of the formula for selecting r objects out of n objects which is $^{n}{{C}_{r}}$.

Complete step-by-step answer:
First, before solving the problem, we try to understand the requirements from the problem. We need to form a committee of 2. Thus, we can have BG, BB or GB OR GG (where B denotes boy and G denotes girl). Now, first we will find the number of outcomes such that the committee of two be formed from three boys and girls. This is given by the formula $^{n}{{C}_{r}}$ (that is for selecting r objects out of n objects). Thus, in this case, we have n = 5 (since total people are there) and r = 2 (selecting a committee of 2 people). Thus, $^{n}{{C}_{r}}$ = $\dfrac{5!}{2!3!}=\dfrac{120}{(2)(6)}$ = 10 (where, n! = $1\times 2\times 3\times ...\times n$ ).
Now, for the committee to contain at least one girl (one or more than one girl in the committee), we make use of few cases –
Case 1: there is one girl in the committee
Outcomes = $^{2}{{C}_{1}}{{\times }^{3}}{{C}_{1}}$ (selecting one girl from two girls and selecting one boy from the 3 boys)
Outcomes = 6
Case 2: there are two girls in the committee
Outcomes = $^{2}{{C}_{2}}$ (selecting two girls from two girls)
Outcomes = 1
Thus, the total number of desired outcomes are 6+1 = 7.
Thus, probability is given by $\dfrac{\text{Desired outcomes}}{\text{Total outcomes}}$ .
Probability = $\dfrac{7}{10}$

Note: Another way to solve the problem is to find the probability that there are no girls in the committee and then subtract the result from 1. This is because –
Probability (no girls in the committee) + Probability (at least one girl in the committee) = 1
For probability for no girls in the committee, the desired number of outcomes are $^{3}{{C}_{2}}$ = 3 (selecting both members from the three boys). Now, the probability is $\dfrac{3}{10}$ (since total outcomes were 10).
Now, we have,
$\dfrac{3}{10}$ + Probability (at least one girl in the committee) = 1
Probability (at least one girl in the committee) = $\dfrac{7}{10}$
Which is the same as obtained in the solution.