Probability of a family having 2 boys and 1 girl

Question

I'm working with the following problem:

Your new neighbors have three children. You are told that they have three children, but without knowing their gender. If you are told about three independent observations of a boy (meaning not knowing whether it's the same boy or different boys), what is the probability that they have two boys and a girl?

Let's form our sample space $S = \{bbb, bbg,....,ggb,ggg\}$

Now we let $H_i = \{$observing $i$ boys$\}$

From this, it becomes evident that $P(H_0) = P(H_3) = 1/8$ whereas $P(H_1) = P(H_2) = 3/8$

We also let $E$ be the event of the family having 2 boys and 1 girl.

From this, we can calculate:

$$P(E|H_0) = P(E|H_3) = 0$$ and $$P(E|H_1) = 3 (1/3)^2 (2/3) = 2/9$$ and lastly $$P(E|H_2) = 3 (2/3)^2 (1/3) = 4/9$$

Now, using Bayes theorem, we get that:

$$P(H_2 | E) = \frac{P(H_2)P(E|A_2)}{P(E)}$$

but $P(E) = P(E|H_1)P(H_1) + P(E|H_2)P(H_2)$ from law of total probability, hence:

$$P(A_2 | E) = \frac{P(H_2)P(E|H_2)}{P(E|H_1)P(H_1) + P(E|H_2)P(H_2)} = $$

$$ = \frac{1}{1+\frac{P(E|H_1)P(H_1)}{P(E|H_2)P(H_2)}} = \frac{1}{1+\frac{2/9}{4/9}} = 2/3$$

but according to my answer sheet, this is not the correct answer. I also can't see where it goes wrong in my solution and would be glad if anyone could help me.

Thanks.

Answer 1

The question is ambiguously worded and assumptions are not clearly stated. If we assume that $(i)$ there are three observations and each observation is an independent event, $(ii)$ in each observation, it is equally likely to see any of the three children, and $(iii)$ each child is equally likely to be a boy or a girl,

Say, the event of observing a boy in all three observations is $A$.

$ \displaystyle \small P(B = 2, G = 1 \mid A) = \frac{P(A \mid B = 2, G = 1) \cdot P(B=2, G =1)}{P(A)}$

$ \small P(A) = P(A \mid B = 3) \cdot P(B=3) + P(A \mid B = 2, G = 1) \cdot P(B=2, G =1) + P(A \mid B = 1, G = 2) \cdot P(B=1, G =3)$

$ \displaystyle \small = \frac 18 + \frac 23 \cdot \frac 23 \cdot \frac 23 \cdot \frac {3}{8} + \frac 13 \cdot \frac 13 \cdot \frac 13 \cdot \frac {3}{8} = \frac 14$

$ \displaystyle \small P(A \mid B = 2, G = 1) \cdot P(B=2, G =1) = \frac 23 \cdot \frac 23 \cdot \frac 23 \cdot \frac {3}{8} = \frac 19$

That leads to,

$ \displaystyle \small P(B = 2, G = 1 \mid A) = \frac 49$

Answer 2

You are not understanding the problem correctly. The probability of observing 3 boys is bigger than $ \frac{1}{8} $. It is bigger than this, since $ \frac{1}{8} $ is the probability that the family does indeed have three boys. If this is the case you are certain to observe three boys. However you can observe three boys even if the family has 1 boy and 2 girls (try to calculate it).
You need to separate (in your head) the probabilities of how many boys the family has and the probabilities of how many boys you are observing.
For starting try to calculate the following:
$ A_i = $ {the family has $i$ boys}
$ H_i = $ {you are observing $i$ boys}
Now what is $ P(H_3|A_i) $, for $ i = 0,1,2,3 $?

Hint: obviously $ P(H_3|A_0) = 0 $ and $ P(H_3|A_3) = 1 $