Question

From a group of 2 boys and 2 girls, two children are selected at random. What is the sample space representing the event? Find the probability that one boy and one girl are selected.

Answer 1

First, we will be defining the terms given in the question that is sample space and the probability. We will first find the number of ways in which 2 children out of 4 is selected by applying the combination formula that is: $^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ and then we will find the number of ways in which 1 boy selected out of the two ways and then the number of ways in which 1 girl is selected out of the two girls. Finally, we will apply the formula for probability formula that is: $\text{Probability of an event to happen}=\text{ }\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

Complete step-by-step solution
Let’s understand what a sample space is. So, a sample space is a collection of a set of possible outcomes of a random experiment. The sample space is represented using the symbol, S. The subset of possible outcomes of an experiment is called events. The sample space for a random experiment is written within curly braces: \left\\{ {} \right\\}.
Let us write the possibilities of selection of two children from the given 2 boys and 2 girls. Let us represent the two boys as B, B’ and represent the girls as G, G’. The possibilities would be: \left\\{ \left( B,G \right),\left( B,G' \right),\left( B',G \right),\left( B'G' \right),\left( BB' \right),\left( GG' \right) \right\\}.
Now, let’s understand what is meant by probability. Probability basically means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one. Now, the probability of all the events in a sample space adds up to 1. The probability formula is defined as the possibility of an event to happen is equal to the ratio of the number of favorable outcomes and the total number of outcomes. Therefore,
$\text{Probability of an event to happen}=\text{ }\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$
Now, we have to find out the probability that one boy and one girl are selected, which means we have to select two children from four. Now, the number of ways that can be done is: $^{4}{{C}_{2}}=\dfrac{4!}{\left( 4-2 \right)!2!}=6$
And, the ways in which 1 boy is selected out of the two boys are: $^{2}{{C}_{1}}=\dfrac{2!}{\left( 2-1 \right)!1!}=2$ , now finally, the ways in which 1 girl is selected out of the two girl are: $^{2}{{C}_{1}}=\dfrac{2!}{\left( 2-1 \right)!1!}=2$.
So, the required probability of selecting one boy and one girl is: $\dfrac{\left( ^{2}{{C}_{1}} \right)\times \left( ^{2}{{C}_{1}} \right)}{^{4}{{C}_{2}}}=\dfrac{2\times 2}{6}=\dfrac{2}{3}$ .
Therefore, the answer is $\dfrac{2}{3}$.

Note: We should know that the sample space of the event is the possibility of selecting two children from the given 2 boys and 2 girls. We may be confused by assuming the sample space of the event as the possibilities of selecting 1 boy and 1 girl from the given 2 boys and girls. We can see that the problem required only a selection of 1 boy and 1 girl, not an arrangement of them. So, we need to use combination not permutation to get the required answer.