1

This is a math problem I came across:

A family has 2 kids. There is an equal probability of (bb, gg, gb, bg). If the first kid is a boy, what is the probability that the other is also a boy?

I tried to solve this using conditional probability.

$$P(\text{other is boy}\mid\text{one is boy}) = \frac{P(\text{other is boy AND one is boy})}{P(\text{one is boy})}.$$

For the numerator, the only case where both conditions are satisfied is (bb):

$$P(\text{other is boy AND one is boy}) = P(\text{bb}) = \frac{1}{4}.$$

For the denominator, “one is boy” includes cases (bb), (bg), and (gb):

$$P(\text{one is boy}) = P(\text{bb}) + P(\text{bg}) + P(\text{gb}) = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}.$$

Thus, the final answer is: $$P(\text{other is boy}\mid\text{one is boy}) = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}.$$

I am having trouble accepting that this is the answer! I really thought the final answer should be 0.5!

I tried to write a computer simulation and got the same answer:

set.seed(123)
n <- 1000000
families <- matrix(sample(c(0, 1), n*2, replace = TRUE), ncol = 2)
at_least_one_boy <- families[families[,1] == 1 | families[,2] == 1, ]
both_boys <- families[families[,1] == 1 & families[,2] == 1, ]
probability <- nrow(both_boys) / nrow(at_least_one_boy)

> probability 
[1] 0.3344364

Is this really correct?

Rócherz
  • 4,241
stats_noob
  • 4,107
  • 2
    You calculated the probability that they have 2 boys given they have at least 1 boy. But you want to order your choices so the denominator should be $1/2$ as there are 2 options where the first child is a boy out of 4 total options. – Nic Mar 06 '25 at 05:03
  • 3
    This is essentially the boy or girl problem. In particular, 1/3 is correct for the question “what fraction of two-child families with at least one boy have two boys?” – Semiclassical Mar 06 '25 at 05:18
  • See https://math.stackexchange.com/questions/1131622/a-very-easy-probability-problem-but-opinions-diverge-greatly-need-some-help and https://math.stackexchange.com/questions/1529796/the-probability-that-the-second-child-of-a-couple-whose-older-child-is-a-boy-is and https://math.stackexchange.com/questions/2769616/two-child-probability-paradox-a-nuanced-explanation and probably a dozen earlier discussions of this problem. Many are listed under Related on this very page. – Gerry Myerson Mar 06 '25 at 05:33

5 Answers5

5

There is no conditionality involved.

Knowing that the first (born) child is a boy doesn't affect the probability of the second child being boy or girl,

Thus P($2nd$ child is a boy) = $\frac12$

$\underline{\texttt{ADDED}}$

Had the wording been one of the two children is a boy, the answer would be different, because the sample space would be $BB, BG, GB$ all equiprobable, giving an answer of $\frac13$

Wording in boy-girl problems has to be precise, with no scope for ambiguity in order to get a correct response.

true blue anil
  • 48,250
1

During the 'For the denominator' part, you took "One child is boy" but according to the question it should actually be "First child is boy"
So the denominator would become $\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$ $$\therefore P(\text{other is boy}\mid\text{one is boy}) = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}$$

Your answer gives the probability for the family having atleast 1 boy and has 2 boys as Nic told in the comments.

Fetray
  • 94
0

First, take the only two cases that are possible since the first child is a boy, which are bb and bg.

Now find out the probability that the second child is also a boy, and the only pair in which this happens in bb. So the probability is $$\frac{1}{1+1}=\frac{1}{2}$$

ThankYouForFlyingRyanair
  • 467
0

Well you want $P(A|B)$ where A is (the second child is a boy) and B is (the first child is a boy)

We know that $P(A|B)=\dfrac{P(A\cap B)}{P(B)}$

Your solution for finding $P(A\cap B)$ is correct but however, $P(B)$ is not the chance that one of them is a boy but the first one is a boy, which is a probability of $\frac{1}{2}$

So you have a total probability of $\dfrac{\frac{1}{4}}{\frac{1}{2}}=\dfrac{1}{2}$.

Alternatively, you can just note that the probability of the events that the first child is a boy and the probability that the second child are independent, so your answer would just be $\frac{1}{2}$.

big poop
  • 1
0

A family has 2 kids. There is an equal probability of (bb, gg, gb, bg). If the first kid is a boy, what is the probability that the other kid is also a boy?

$$P(\text{other is boy}\mid\text{one is boy}) = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}.$$ Is this really correct? I really thought the final answer should be $0.5$!

Yes, the correct answer is indeed $\dfrac13:$ being given that the first pick is male reduces the sample space by eliminiating the outcome 'gg', so the second pick's probability of being male is no longer $\dfrac12$ as if the sample space was still 'bb, gg, gb, bg'!

Distilling the question without loss of generality:

  • $bb, gg, gb, bg$ have equal probabilities. If the first kid is a boy one kid's gender is known, what is the probability that the other kid is also a boy has the same gender?

This makes clearer that the question is really asking for the probability that the two picks have the same gender given only the gender, rather than the probability of the second pick independently being male.

ryang
  • 44,428