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A couple plans on having 2 children. Given that at least one of them is a boy, the probability that both are boys is

$$ \frac{P(both~boys)}{P(at~least~one~is~a~boy)} = \frac{0.25}{0.75} =\frac{1}{3} $$

Furthermore, a textbook I am reading claims that the probability of both children being boys given that at least one is a boy born in spring is $\frac{7}{15}$. It doesn't explain its solution.

Why? What does the season have anything to do with whether both children are boys?

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    $\begingroup$ Obviously, $\frac7{15}$ stands for the ratio of $1-\left(\frac34\right)^2$ by $1-\left(\frac14\right)^2$. Why these appear is not clear to me. Which textbook is claiming this? $\endgroup$
    – Did
    Commented Sep 18, 2012 at 20:44
  • $\begingroup$ I think spring has to do with the cartezian product of the sample space that we had for the previous case, where we were interested in 2 children $\Omega_0=\{00,01,10,11\}$ and the seasons $\Omega_1=\{00,01,10,11\}$, with winter=00, autumn=01, spring=10 and summer=11. Finally what we are talking about is over $\Omega=\Omega_0 \times \Omega_1$. $\endgroup$
    – Seyhmus Güngören
    Commented Sep 18, 2012 at 21:20
  • $\begingroup$ Sample space should have $64$ elements, unlike what I wrote previously as a comment. There are two cases "first boy-second boy" each having $8$ different possibilities, such as \{boy and born in spring, boy and born in autumn,...,girl born in summer\}. In this case I find @Adam ' s answer more systematic (+). $\endgroup$
    – Seyhmus Güngören
    Commented Sep 18, 2012 at 22:44
  • $\begingroup$ @Did 1. Blitzstein, Introduction to Probability (2019 2 ed) Example 2.2.7, p 51 has this same question, but features "winter" rather than "spring". But which season is irrelevant, because a different season doesn't change the answer to this question. 2. See the bottom of web.archive.org/web/20210516174259/http://adit.io/posts/…. $\endgroup$
    – user53259
    Commented Jul 26, 2021 at 23:37

4 Answers 4

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Specifying one of the boys was born in spring, increases the probability the second child is also a boy, because parents with two boys are more likely to have one born in spring than parents with just one. You are basically calculating:

$$ \frac{P_A}{P_B}=\frac{\frac{1}{4}(P_D+P_E)}{1-P_C}=\frac{\frac{1}{4}(\frac{6}{16}+\frac{1}{16})}{1-(\frac{7}{8})^2}=\frac{7}{15} $$ With $P_A$ as the probability of having two boys, at least one of which born in spring; $P_B$ of having at least one boy born in spring; $P_C$ of having no boys born in spring; $P_D$ of exactly one out of two children being born in spring; and $P_E$ of two out of two children being born in spring.

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  • $\begingroup$ It actually decreases the probability that the second child is a boy (because in the case if both boys being born in spring, ordering will not matter, resulting in one less case for the sample space). $\endgroup$
    – rrampage
    Commented Sep 18, 2012 at 21:46
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    $\begingroup$ Err... 7/15 > 1/3 $\endgroup$
    – Marcks Thomas
    Commented Sep 18, 2012 at 22:25
  • $\begingroup$ Oops.. I was thinking in terms on 8/15 > 7/15. Sorry for the scramble. :P $\endgroup$
    – rrampage
    Commented Sep 18, 2012 at 22:37
  • $\begingroup$ Thanks. Where did the numerator $\frac{1}{4}(\frac{6}{16}+\frac{1}{16})$ come from? Why is $\frac{1}{4}$ being multiplied to the latter part? Isn't the latter part enough? $\endgroup$
    – Minden Petrofsky
    Commented Sep 19, 2012 at 2:35
  • $\begingroup$ Probability of having two boys (1/4), multiplied by the probability of having at least one of them born in spring. That's $\frac{3}{16} + \frac{3}{16} + \frac{1}{16}$, or alternatively: $1-(\frac{3}{4})^2$. $\endgroup$
    – Marcks Thomas
    Commented Sep 19, 2012 at 10:19
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If the probability that a child is a boy is $\frac 12$ and the probability that it is born in spring is $\frac 14$, the probability that it is a boy born in spring is $\frac 18$. The probability that at least one of two children is a boy born in spring is therefore $1 - (1 - \frac 18)^2$ = $\frac{15}{64}$.

The probability that both children are boys born in spring is $(\frac 18)^2 = \frac{1}{64}$, so the probability that exactly one boy is born in spring is $\frac{15-1}{64}$. If exactly one boy is born in spring, the probability that the other child is a boy (not a boy born in spring, so either a boy born in one of the other three seasons or a girl born in any of the four seasons) is $\frac{3}{3+4}=\frac{3}{7}$. The probability that one child is a boy born in spring and the other is a boy born in one of the other three seasons is therefore $\frac{15-1}{64}$ x $ \frac{3}{7}$ = $\frac{6}{64}$. This makes a total joint probability of two boys and at least one boy born in spring of $\frac{1+6}{64}=\frac{7}{64}$.

The desired conditional probability is therefore $\frac{7}{64}$ / $\frac{15}{64}$ = $\frac{7}{15}$.

Addendum 24/9/2021 Addressing questions in comments by koss:

Q1 “Why does Pr(exactly 1 boy born in spring) = Pr(At least 1 of the 2 children is a boy born in spring) – Pr(both children are boys born in spring)?”

A1 The following three possibilities are exhaustive: a) No boys born in spring; b) Exactly 1 boy born in spring; c) 2 boys born in spring. Pr(At least 1 is a boy born in spring) = Pr(b) + Pr(c). Therefore Pr(b) = Pr(At least 1 is a boy born in spring) – Pr(c).

Q2 “Why must you add 1/64 to 6/64?”

A2 6/64 is the probability that one child is a boy born in spring and the other is a boy born in one of the other three seasons. Thus it excludes the probability that both boys were born in spring. But we are also interested in the latter probability because the question asks for the probability of both children being boys given that at least one is a boy born in spring. So we must add that probability which is 1/64.

Q3 “Why isn’t Pr(one child is a boy born in spring and the other is a boy born in one of the other three seasons) = Pr(two boys and at least one boy born in spring)?

A3 The former excludes, but the latter includes, the case of two boys both born in spring.

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  • $\begingroup$ you get boy born in spring and the other boy born in another season is $14/64\times 3/7$. It is easier to say : since both are the boys I have probability $16/64$. There are $16$ cases out of $64$ but one boy should have been born in spring and the other in other $3$ moths =$3$ cases now the second boy was born in spring and the first boy in another season $3$ more case, altogether $6$ out of $16$ and finally adding upto $6/64$ $\endgroup$
    – Seyhmus Güngören
    Commented Sep 18, 2012 at 23:15
  • $\begingroup$ Thanks Adam! "The probability that the other child is a boy (not a boy born in spring, so either a boy born in one of the other three seasons or a girl born in any of the four seasons) is $\frac37$." Can you please elaborate this? How did you deduce $\frac37$? What does 3 mean? What does 7 mean? $\endgroup$
    – user851668
    Commented Sep 1, 2021 at 8:16
  • $\begingroup$ Can you please respond in, by editing, your answer? Comment chains are cumbersome to read. $\endgroup$
    – user851668
    Commented Sep 1, 2021 at 8:16
  • $\begingroup$ @Intellectuallydisabled Edited as requested. $\endgroup$
    – Adam Bailey
    Commented Sep 2, 2021 at 9:30
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A couple plans on having 2 children. Judging from the probabilities 0.25 of both boys and 0.75 of at least 1 boy, I can surely say that P(boy) = 0.5 and P(girl)=0.5.

I presume that you are asking how is the probability of both boys given that at least 1 boy is born in spring = $\frac{7}{15}$

Now, since the word "spring" is somehow introduced, lets assume that it is possible for a child to be born in any of the 4 seasons with equal probability.

Lets now have a look at the sample space:

A = All events in Set(Both Boys | One is born in Spring) = 2*(Boys in Autumn, Winter, Summer) + (Both boys in Spring)

B = All events in Set (One Boy born in Spring) = 2*(Boys in Autumn, Winter, Summer) + (Both boys in Spring) + 2*(Girl in Autumn, Winter, Summer, Spring)

Hence, Probability of (Both Boys|One boy is born in spring) = $\frac{P(A)}{P(B)}$

We assume no chance of twins.

Hope that solves the question.

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A simple way is to keep track of the older child.

If the older child is a boy born in spring, the younger child could be a girl or boy born in any season ($8$ cases, $4$ of them favorable)

If the older child is not a boy born in spring, the younger child is. The older child could then be a girl born in any season, or a boy born in non-spring
($7$ cases, $3$ of them favorable)

Thus $Pr = \Large{\frac{4+3}{{8+7}} = \frac{7}{15}}$

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