Question

In a family of 3 children, the probability of having at least one boy is
A. $\dfrac{7}{8}$
B. $\dfrac{1}{8}$
C. $\dfrac{5}{8}$
D. $\dfrac{3}{1}$

Answer 1

Hint: Probability is a fraction of favorable cases for an event and total cases. “At last one boy” means there may be one boy or two boy or three boy of 3 children in the family. Find the probability of having no boy of 3 children and subtract it from 1 to get the answer.

Complete step-by-step answer:
Here, we need to find the probability of having at least one boy out of 3 children in a family. So, as we know that the significance of at least one boy of 3 children is that there must be one boy of 3 children or two boys of 3 children or 3 boys of 3 children, so we need to find the probability of having one boy or 2 boy or 3 boy out of 3 children in the family.
As we know the probability can be defined mathematically as the ratio of favorable number of cases to the total number of cases for the given event.
And we know the property of probability of an event that
$P\left( E \right)+P\left( {{E}^{1}} \right)=1$ …………. (i)
Where P (E) is the probability of occurring of the event and P(E) is the probability of occurring of the event and $P\left( {{E}^{1}} \right)$is the probability of not occurring of the event E.
Now, coming to the question, we can write the equation using the equation (i) as
$\Rightarrow $ Probability of getting at least one boy + probability of getting no boy = 1
Or
Probability of getting at least one boy = 1 – probability of getting no boy ……………….. (ii)
We know that the total number of cases for the 3 children to be boy or girl are given as:
(i) 0 boy 3 girl
(ii) 1 boy 2 girl
(iii) 2 boy 1 girl
(iv) 3 boy 0 girl
As there are half chances of having a boy or girl. It means the probability of having a boy or girl is $\dfrac{1}{2}$.
Hence, we get
$P\left( G \right)=P\left( B \right)=\dfrac{1}{2}$ ……………….. (iii)
Where P(G) and P(B) are the probabilities of having a boy and a girl.
It means probability of having no boy out of 3 children is given as
P = probability of getting 3 girls out of 3 children
P = P(B) . P(B) . P(B)
$P=\dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}$
$P=\dfrac{1}{8}$………………… (iv)
Now, put this probability of getting no boy in the equation (ii) as
Probability of getting at least one boy
 $\begin{align}
  & =1-\dfrac{1}{8} \\
 & =\dfrac{8-1}{8}=\dfrac{7}{8} \\
\end{align}$
Hence, the probability of having at least one boy is $\dfrac{7}{8}$.
So, option (a) is correct.

Note: Another approach for the question would be that we can add probabilities of having 1 boy, 2 boy, 3 boy of 3 children in the family as:
We need to use binomial relation i.e. $C_{r}^{n}{{p}^{r}}{{q}^{n-r}}$ where P is the probability of success and q is the probability of failure.
So, take P = success = having a boy = $\dfrac{1}{2}$
q = failure = having a girl = $\dfrac{1}{2}$
$P=C_{1}^{3}{{\left( \dfrac{1}{2} \right)}^{1}}{{\left( \dfrac{1}{2} \right)}^{2}}+C_{2}^{3}{{\left( \dfrac{1}{2} \right)}^{2}}{{\left( \dfrac{1}{2} \right)}^{1}}+C_{3}^{3}{{\left( \dfrac{1}{2} \right)}^{3}}$
$P=\dfrac{3}{8}+\dfrac{3}{8}+\dfrac{1}{8}=\dfrac{7}{8}$
So, it can be another approach for the question.
Understanding the term ‘at least one boy’ and writing it as difference 1 and probability of having no boy is the key point of the question and provided solution.