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Suppose a family has two children. The probability of a child being a boy is $1/2$, independent of the other. If we are given that one child is a boy, what is the probability the other is a boy as well.

I would calculate the probability as follows:

$$P(\text{two boys | at least one boy)} = \dfrac{P(\text{two boys)}}{P(\text{at least one boy})}.$$

Now, the probability of having two boys is clearly $1/4$. The bit that troubles me is the probability of having one boy. I see two ways to tackle this:

  1. Suppose we count the children as (child 1, child 2). Then there are four possibilities: (boy, girl), (girl, boy), (boy, boy), (girl, girl). Then $P(\text{at least one boy})= 3/4$.

  2. On the other hand, suppose I count the children not in order, but like {child 1, child 2}. Then there are $4-1$ possibilities: {boy, girl} = {girl, boy}, {boy, boy}, {girl, girl}. Hence $P(\text{at least one boy})= 2/3$

These give different answers... What should I do?

N. F. Taussig
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Stijn D'hondt
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    2/3 is correct because the case of two girls is ruled out, the remaining three possibilities have the same likelihood. – Peter Feb 15 '20 at 11:06
  • In your second scenario, your three cases are not equally likely, so your calculation there is incorrect. – B. Goddard Feb 15 '20 at 11:06
  • Probability of "at least one boy" is 3/4, so your calculation also gives 1/3 for two boys. – Peter Feb 15 '20 at 11:10
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    Peter and Goddard are contradicting each other... And why would P(two boys) be 1/3, Peter? It should definitely be 1/4, since P(boy) = 1/2. – Stijn D'hondt Feb 15 '20 at 11:14
  • This is a conditional probability – Peter Feb 15 '20 at 11:20
  • Are we given that "at least one is a boy" or that "this one is a boy"? If you are only told the first, then you would be less certain that both are boys, because you would have to check both to find out (the probability is $1/3$). Where as if you were told a specific one was a boy, you only need check the other to verify if both are (and so the probability is $1/2$). – Graham Kemp Feb 15 '20 at 11:30
  • @PaleBlueDot We're not contradicting each other. Peter is correct and calculates the probability from the first scenario. I'm asserting that the second scenario is incorrect. It would give the final answer as (1/4)/(2/3) = 3/8. – B. Goddard Feb 15 '20 at 11:36
  • Related: https://math.stackexchange.com/questions/15055/in-a-family-with-two-children-what-are-the-chances-if-one-of-the-children-is-a – Henry Jun 01 '20 at 09:36

2 Answers2

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Order is important, you can also relate two child as elder child and younger child, so lets call it {elder, younger}. Now, the problem we are facing is weather to consider {boy,girl} and {girl,boy} as same or not. Off-course if we consider it as an ordered pair so they are different. If we would have considered it as just a collection or set then they are same.

Also it is important that the events of sample space to be equally likely. So, Considering as ordered pair makes the events of sample space equally likely, where as if we would have considered as set or collection than the {boy, girl} as set would be twice as likely to happen then the other events in sample space.

Or u can see it as, if the first child is boy/girl still it can fall in {boy, girl} set, and only after the child 2 it is in {boy, girl} set with 50% chance. For other sets like {boy, boy} the chance if 25%.

So it is about sample space with equally likely events.

Shreyas Bhatt
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For visualization, you are assuming that 'the event has been made' and there are 2 people with boy and girl being equally likely. Then, the simple equations hold: $$P(\text{two boys}) = 1/4$$ and $$P(\text{at least one boy})= 3/4$$.

Now it is revealed that one of those is a boy, but you don't know which one. This leaves you with three out of four possibilities: (boy, girl), (girl, boy), (boy, boy).

Only one of them is "desirable", so the chance is 1/3, which could also be computed as you suggested: $$\dfrac{P(\text{two boys)}}{P(\text{at least one boy})} = 1/3$$.

Your 2nd option doesn't hold, since you merged two different outcomes into one (as we said, we don't know which one is a boy). (If you knew which one is a boy, then you'd only care about the other one being a boy and the probability would simply be $1/2$.)

h4nek
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