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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?

  • 0
    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 (+).2012-09-18

4 Answers 4

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

6

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.

  • 0
    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: $1-(\frac{3}{4})^2$.2012-09-19
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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 $(\frac 18)^2 = \frac{1}{64}$. 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 $\frac{15-1}{64}$ x $ \frac37$ = $\frac{6}{64}$ (multiplying by $\frac37$ rather than $\frac48$ because $\frac{15-1}{64}$ excludes the case in which the other child is also a boy born in spring). This makes a total joint probability of two boys and at least one boy born in spring of $\frac{7}{64}$.

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

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    Have edited to explain the 3/7 which I realise was not entirely clear.2012-09-19
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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.