Question

In a family of $3$ children, find the probability of having at least one boy.

Answer 1

Hint: In this question we will use the concept of probability. In this question first we will make some events according to the statements given in the question and we will make the sample space of the different case and then by using the sample space we will find out the required probability.

Complete step-by-step solution -
Given that , the family have $3$ children,
We know that there will be only two possibilities ,that either the child would be a boy or a girl.
Now, for $3$ children ,we will make a sample space for different cases,
( We know that , in a sample space there will be ${2^n}$ sample points so here, the value of $n = 3$ then the number of sample points in this sample space =${2^3} = 8$ )
Sample space = [ $bbb,bbg,bgb,bgg,gbb,gbg,ggb,ggg$ ]
( Here $b = boy{\text{ and }}g = girl$)
Now we have to find the probability of having at least one boy ,
So from the sample we can see those cases in which have at least one boy
Let E be the event having at least one boy ,
$E = [{\text{ }}bbb,bbg,bgb,bgg,gbb,gbg,ggb{\text{ ]}}$
We know that the probability of an event =$\dfrac{{{\text{number of favourable outcomes}}}}{{{\text{total possible outcomes }}}}$ .
So, the probability of having at least one boy = $\dfrac{7}{8}$.

Note: In this type of question first we will make the sample space of the possible outcomes and then by finding out the favourable cases according to the question we will find out the probability of the given event by using the formula of probability. we can also solve this question by finding out the probability of having no boy ,and then by using the formula of probability of event (not A) i.e. $P(notA) = 1 - P(A)$, We will find out the probability of at least one boy.