Question

If at least one child in a family with 3 children is a boy then the probability that 2 of the children are boys, is
A: $\dfrac{3}{7}$
B: $\dfrac{4}{7}$
C: $\dfrac{1}{3}$
D: $\dfrac{3}{8}$

Answer 1

Hint: First, try to find all the possible ways, as they have given in the question find the possible ways with at least $1$ boy. Then out of all the possible ways with at least one boy, find how many ways having two boy children will give us the correct answer.

Complete step-by-step answer:
First we will understand the question they have given,
They have given that in a family there are $3$ children’s. So now we will write all the possible ways of having a girl and boy child.
We will take g for girl child and b for boy child and see the possible ways,
1: g g g
2: g g b
3: g b b
4: b b b
5: b b g
6: b g g
7: b g b
8: g b g
We have written all the possible ways for $3$ children but in the question they have given that at least $1$ boy should be present so case 1 with all the girls is excluded.
Now, the total number of ways with at least one boy is present will be $7$.
Now, we need to find out the total number of ways which is having $2$ of the children are boys
From the above listed possibility we can see that in case $1,5,7$ are having $2$boys and $1$ girl.
So the probability of $2$ of the children are boys, can be written by using the probability formula that is,
$P(A) = \dfrac{{n(A)}}{n}$
Where $P(A)$ denotes the probability of A
$n(A)$ is the number of occurrence of A or the number of favorable outcomes, here it is $3$
$n$ is the total number of possible outcomes or the sample space, here it is $7$.
Hence, we can write the probability as $P(A) = \dfrac{3}{7}$.
Therefore the correct answer is option A.

Note: In the question they have mentioned out of $3$ children’s one should be boy so while considering the sample space you should not consider the possibility of having all the girls, by excluding the one possibility we need to consider the other for sample space.