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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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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? –  Did Sep 18 '12 at 20:44
    
Obviously...... –  Graphth Sep 18 '12 at 21:14
    
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 $\Omega=\Omega_0 \times \Omega_1$. –  Seyhmus Güngören Sep 18 '12 at 21:20
    
@did posted what I think is the right answer. Have a look. –  rrampage Sep 18 '12 at 21:29
    
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 (+). –  Seyhmus Güngören Sep 18 '12 at 22:44
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4 Answers 4

up vote 4 down vote accepted

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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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 $6/64$ –  Seyhmus Güngören Sep 18 '12 at 23:15
    
Thank you, where did $\frac{3}{7}$ come from? –  Minden Petrofsky Sep 19 '12 at 2:26
    
Have edited to explain the 3/7 which I realise was not entirely clear. –  Adam Bailey Sep 19 '12 at 8:45
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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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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). –  rrampage Sep 18 '12 at 21:46
    
Err... 7/15 > 1/3 –  Marcks Thomas Sep 18 '12 at 22:25
    
Oops.. I was thinking in terms on 8/15 > 7/15. Sorry for the scramble. :P –  rrampage Sep 18 '12 at 22:37
    
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? –  Minden Petrofsky Sep 19 '12 at 2:35
    
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: $1-(\frac{3}{4})^2$. –  Marcks Thomas Sep 19 '12 at 10:19
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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.

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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)

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)

Indicated Pr = 7/15

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