# Sample space: What's the possibility that a family has $n$ boys?

What's the sample space in these two cases?

Case $$1$$:
Of all families with two children you ask the parents, if they have a boy born on a Thursday. They say yes. What's the possibility that the family has two boys?
I'd say the sample space $$\Omega=\{BG, GB, BB\}$$ (because we know there is already a boy).
So the probability would be $$1/3.$$

Case $$2$$:
You know a family has three kids. On a day your visiting them, you see two boys playing in the garden. What's the possibility the family has three boys?
I'd say the sample space $$\Omega=\{GBB, BGB, BBG, BBB\}$$ so the probability would be $$1/4.$$
Or is the sample space $$\Omega=\{B, G\}$$, because you already know, they have two boys? Which would be a probability of $$1/2.$$

Thanks in advance

• For problems like these (especially case 1), I don't think there is any such thing as "the" sample space in general. There are only correct sample spaces and incorrect sample spaces. I think you can also reasonably define the sample space to be the set of all possible family configurations before you make observe any of the children, or after, as long as you apply probabilities correctly to the sample space you use. (If you use the "before" sample space then you need to use conditional probability.) Jul 1, 2015 at 13:43

## 3 Answers

Case 1 is badly worded. Do you mean you deal with families with two children, and you ask the probability that such a family has two boys given that they have a boy born on a Thursday? And do you want a sample space where the outcomes have equal probability?

If so, let the days of the week be $SMTWHFs$. Then your sample space begins with $BSBS$, meaning the first child was a Boy born on Sunday and the second child was a Boy born on Sunday. This sample space has $2\cdot 7\cdot 2\cdot 7=196$ outcomes. You want the conditional probability of two $B$'s given at least one $BH$.

There are other ways to do this with smaller sample spaces, if you allow the sample space to be non-uniform in probability.

Let's work it out with my sample space. We want

$$P(2\text{ boys | a boy born Thursday})=\frac{P(2\text{ boys and a boy born Thursday})}{P(\text{a boy born Thursday})}$$

For the numerator, $2$ boys and a boy born Thursday, we have three overlapping subcases: the first boy born Thursday ($7$ possibilities for the other child); the second boy born Thursday ($7$ possibilities for the other child); or both boys born Thursday ($1$ possibility). The third case is the double counting if we add the first two cases, so the total number of possibilities is

$$P(2\text{ boys and a boy born Thursday})=7+7-1=13$$

For the denominator, we again have three overlapping subcases: the first child is a boy born Thursday ($2\cdot 7$ possibilities [sex and day] for the other child); the second child is a boy born Thursday ($2\cdot 7$ possibilities [sex and day] for the other child); or both children are boys born Thursday ($1$ possibility). Again, the third case is the double counting if we add the first two cases, so the total number of possibilities is

$$P(\text{a boy born Thursday})=2\cdot 7+2\cdot 7-1=27$$

Therefore we get

$$P(2\text{ boys | a boy born Thursday})=\frac{P(2\text{ boys and a boy born Thursday})}{P(\text{a boy born Thursday})}=\frac{13}{27}$$

This may seem strange, but the information about a birth on Thursday is actually meaningful to the problem, since it restricts both numerator and denominator in the calculation but in different proportions. For more details, see this article or this article on the problem where "Thursday" is replaced by "Tuesday." I just tested this problem in a Microsoft Excel spreadsheet, listing all $196$ outcomes, and it checks out.

Your rewording of the problem is more clear, and it seems you do not want the sample space to be all families with two children but rather those with two children with at least one of them a boy born on Thursday. That does not change my answer but does change the overall sample space. In this situation your sample space has $27$ elements, namely

1st Boy-Thursday  2nd Boy-Thursday
BHBS              BSBH
BHBM              BMBH
BHBT              BTBH
BHBW              BWBH
BHBH
BHBF              BFBH
BHBs              BsBH
BHGS              GSBH
BHGM              GMBH
BHGT              GTBH
BHGW              GWBH
BHGH              GHBH
BHGF              GFBH
BHGs              GsBH


Note that the second column has a gap, to avoid double-counting the case where both children are boys born on Thursday. You can easily count that $13$ of these $27$ possibilities have two boys.

Case 2 is more clearly worded. Your sample space is $BBB$ through $GGG$, a size of $8$. You want the conditional probability that there are three boys given that there are at least two boys. Your calculation here of $\frac 14$ is correct.

• Thanks, but which calculation is correct in case 2? The one with sample space Omega={GBB, BBG; BGB, BBB} or Omega={B, G} Jul 1, 2015 at 8:59
• I updated case 1. Because we know they have a boy on Thursday. I wrote that wrong. Sorry. Jul 1, 2015 at 9:02
• @mblaettler: In case 2 your calculation of $\frac 14$ is correct because you used the correct sample space of size $8$ with equal probabilities. You did the conditional probability by looking at the event "has at least two boys" of size $4$. Your second calculation of $\frac 12$ is wrong because the two listed outcomes do not have the same probability. Jul 1, 2015 at 10:07

I will deal with the first case. Consider two subcases; first assume you have seen the oldest boy. Then, the younger child is either a boy or a girl, both with probability $0.5$. Hence the probability of the family having two boys equals $0.5$.

Conversely, if you assume you have seen the youngest boy, then the older child is either a boy or a girl with probability $0.5$, which yields the same argument. In any case, the probability that the family has two boys equals $0.5$.

This same reasoning can be applied in a more elaborate manner to get to a probability of $0.5$ in the second case as well.

The reason why there is confusion is because the sample space $\Omega$ is not the whole story; there is also a probability measure $\mathbb{P}$ that assigns a probability to each event in $\Omega$. These events are not uniformly distributed given the information that the family has at least one boy.

• Thanks, but in case 1, we didn't know it's the older boy. We just know the family has one boy, born on Thursday. So there are three possibilities left, aren't there? Jul 1, 2015 at 8:55
• First I assume we see the older child (which happens to be a boy), and then I assume we see the younger child. In both cases the probability of two boys is $0.5$. Jul 1, 2015 at 8:55

Case 2:

I'll deal with this case first, as it is simpler.

Essentially you are asked P(3 boys | at least 2 boys)

The sample space Ω={GBB, BGB, BBG, BBB} given by you is correct.

Case 1:

I take it that the family has two children of which a boy (= at least one boy) was born on Thursday.

The sample space is thus Ω={GB, BG, BB}

Important Edit

It appears that a son being born on Thursday has a probabilistic impact. Rather than my trying to explain, see a similar problem here

The indicated probability turns out to be $\frac{13}{27}$, and (by the same approach) if it is meant that exactly 1 boy was born on Thursday, $\frac{12}{26} = \frac{6}{13}$