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How do I find the probability of a three children family having exactly two boys given that at least one of their children is a boy?

Do I use the dependent formula $$P(A \text{ and } B) = P(A) \times P(B \text{ given that }A \text{ has occurred})$$ or do I use the conditional probability form of $P(B|A)$?

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  • $\begingroup$ The first works just fine. $\endgroup$ May 17, 2012 at 19:00
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    $\begingroup$ Given that indicates conditional probability. $\endgroup$
    – Did
    May 17, 2012 at 19:00
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    $\begingroup$ If you haven't internalized the concepts it will be easier to build intuition if you enumerate the sample space and count the desired outcomes. E.g. at least one boy excludes GGG from the unrestricted set of eight {BBB, BBG, ... }. $\endgroup$
    – karakfa
    May 17, 2012 at 19:00

2 Answers 2

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$\Pr[B\mid A]$ is the same as $\Pr[\text{$B$ given that $A$ has occurred}]$. Therefore, if you divide both sides of $\Pr[\text{$A$ and $B$}]=\Pr[A]\cdot\Pr[\text{$B$ given that $A$ has occurred}]$ by $\Pr[A]$, you get $$ \Pr[B\mid A]=\frac{\Pr[\text{$A$ and $B$}]}{\Pr[A]}, $$ which is the conditional probability formula.

This can be used to solve your problem. Write $$ \begin{aligned} \Pr[\text{exactly 2 boys}\mid\text{at least 1 boy}]&=\frac{\Pr[\text{exactly 2 boys and at least 1 boy}]}{\Pr[\text{at least 1 boy}]}\\ &=\frac{\Pr[\text{exactly 2 boys}]}{\Pr[\text{at least 1 boy}]}. \end{aligned} $$ The second line follows from the first because there being exactly two boys implies that there is at least one boy.

If you treat the situation as a Bernoulli process, you can compute both of the needed probabilities.

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given that ,at least one children is boy,consider different situation .first consider that we have information about one boy,then probability that in remained two children one of them is boy is $1/2=0.5$ if it has two boys then probability is $1$,but if three boy,then probability that they have exactly two boy,is zero,i think it would help you

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