GATE DS&AI 2024 | GA Question: 7

Question

The probability of a boy or a girl being born is $1 / 2$. For a family having only three children, what is the probability of having two girls and one boy?

• A. $3 / 8$
• B. $1 / 8$
• C. $1 / 4$
• D. $1 / 2$

Answer 1

Ans I think would be A. 3/8

As given, Probability of a boy or girl being born is 1/2.
Also it is asked that what is the probability of having 2 girls and 1 boy for a family having ONLY 3 children.

Therefore,
We have a combination of

= B*G*G + G*B*G + G*G*B
= 1/2 * 1/2 * 1/2 + 1/2 * 1/2 * 1/2 + 1/2 * 1/2 * 1/2
= 3/8.

Answer 2

Given that there are 3 children, the possibilities are:

1. All boys - B B B

2. All girls - G G G

3. 2 boys and 1 girl - B B G

4. 2 girls and 1 boy - G G B

Observe that here the order of children being born would not matter, as any ordering will contain the same number of boys/girls.

So the required probability would be 1/4.

Comments

I support this answer

---

when we define the probability, we also have to define the sample space. In your case sample space is having 4 events and your answer indicates that they are equally likely.

It means that having 3 boys also have the probability 1/4 but as per the given question it should be 1/2*1/2*1/2=1/8 assuming having any boy is an independent event.

Order does not matter that's why answer is $\frac{\#\{GGB,GBG,BGG\}}{\#\{BBB,BBG,BGB,BGG,GBB,BBG,GGB,GGG\}}=\frac{3}{8}$

Answer 3

A. 3/8

There are mutually exclusive and exhaustive cases.

P(all 3 child boy) + P(all 3 child girl ) + P(2 boy and 1 girl child ) + P(1 boy and 2 girl child) =1.

Because of symmetry cases, P(all 3 child boy)  = P(all 3 child girl ) =1/8

so, P(2 boy and 1 girl child ) = P(1 boy and 2 girl child) = (1-1/8 -1/8)/2 =3/8

Answer 4

3 children where each of them can be either B or G - so $2^3 = 8$ possibilities.

Favorable case is any permutation of "BGG". So, number of favorable cases $ = 3!/2! = 3.$

So, required probability $ = 3/8.$