The question is:

A certain family has 6 children, consisting of 3 boys and 3 girls. Assuming that

all birth orders are equally likely, what is the probability that the 3 eldest

children are the 3 girls?

I thought it would just be $(\frac{1}{2})^6$ since the probability of the first person being a girl is $\frac{1}{2}$, the second person being a girl is $\frac{1}{2}$, etc. Does anyone know why this thinking is incorrect?

  • 1
    $\begingroup$ We are told there are $3$ boys. So the results are very far from independent. You calculated the probability there are $3$ girls followed by $3$ boys, given no information. $\endgroup$ – André Nicolas May 19 '16 at 22:03
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    $\begingroup$ If you have a bucket with three pink and three blue balls in it and you draw balls from the bucket independently and uniformly at random without replacement, what is the probability when drawing three balls that you draw specifically three pink balls? What was the chance of the first ball being pink? Given that the first ball is pink what is the chance of the second ball being pink also? (there are fewer pink balls in the bucket than blue now) Given that the first two are pink, what is probability of final being pink? Apply multiplication principle. Relate your problem to this. $\endgroup$ – JMoravitz May 19 '16 at 22:09

In your calculation you didn't account for the fact that we already know that there are $3$ boys and $3$ girls. Probabilities are the ratios of the events we are interested in over all possible events. In the calculation you made, it implicitly assumes that the family could have all boys or all girls. However, we are given that the family exactly $3$ girls and $3$ boys, which needs to be accounted for in the calculation.

Here is how it can be done:

In the family you have $6$ children. You are interested in how many ways exist where exactly $3$ are girls (or equivalently $3$ are boys). So you are interested in $1$ out of all possible arrangements with exactly $3$ girls. To find the total number of such arrangements we look for how many ways we can choose $3$ out of $6$, which is given by $\binom{6}{3}=20$. Thus the probability of the three eldest being girls is $$\frac{1}{\binom{6}{3}}=\frac{1}{ 20}.$$


The given condition in the problem is that the family has three boys and three girls. The possibility of a family having 3 boys and 3 girls is actually $\begin{pmatrix}6\\3\end{pmatrix}/64=\frac{20}{64}$. Then you can use your $\frac{1}{64}$ to do a conditional probability $\frac{1}{64}/\frac{20}{64}=\frac{1}{20}=0.05$

We can use combination to solve this problem. There are $\begin{pmatrix}6\\3\end{pmatrix}$ways to choose where the girls appear in the birth order (no order)
Of these possibilities, there is only 1 way where the 3 girls are the 3 eldest children. So the probability is $1/\begin{pmatrix}6\\3\end{pmatrix}=0.05$

Permutation also works here. Label the girls as 123 and the boys as 456. Think of the birth order is a permutation of 123456. The number of possible permutations of the birth orders is 6!. Now count how many of these have 123 appear before 456. We have $\frac{(3!)(3!)}{6!} = 0.05$.


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