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I had this question on an exam last week: there is a family, and they have $3$ children, you know that they have exactly $2$ boys and $1$ girl. What is the probability that the first born is a boy?

Some people answered $1/2$ and others $2/3$, what do you think? We can't figure it out.

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We can use Bayes' Theorem for conditional probability problems:

$$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$$

In this case, the events A and B are:

A: first born is boy

B: family have exactly two boys, one girl

Bayes' Theorem then states for this problem that:

The probability of the first born being a boy given that the family has two boys and a girl is:

the probability that the family has two boys and a girl given the first born is male, times the probability the first born is a boy, divided by the probability the family has two boys and a girl.

Each of the probabilities needed can be calculated:

$$ \begin{align} P(B|A)&=P(\text{the family has two boys and a girl|the first born is male})=\frac12\\ P(A)&=P(\text{first born is a boy})=\frac12\\ P(B)&=P(\text{the family has two boys and a girl})=\frac38 \end{align} $$

Putting these together gives:

$$P(A|B)=\frac{\frac12\frac12}{\frac38}=\frac23$$

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Possibilities are: bbg bgb gbb

Favorable: bbg bgb

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The three different possibilities are BGB - BBG - GBB => Probability of a boy first is two thirds.

Another way to look at it: call the children X,Y,Z, with Y and Z males, and X female. So what is the probability that X is first? It's 1/3, since they are all equally likely to come up first. So with probability 2/3 the first one is either Y or Z, i.e. a male.

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  • $\begingroup$ Thank you! This will clear out a lot of discussions! $\endgroup$ – ZeniaA May 20 '17 at 7:50

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