# If a random family is chosen and then a random child is chosen from the family, what is the probability that the child is female?

I am having some difficulty with this question and any help would be appreciated! I have managed to solve the first two parts, but I am not sure how to express the effect of 'choosing a random family' on the sample space.

For example, how does part (d) differ from simply finding $P(N=2|F)$ (which was already asked in an earlier part?

A certain country town has $4$ families having respectively $4, 3, 2, 2$ children with genders FFMM, FFM, FF, FM.

$(c)$ If a random family is chosen from the town and then a random child from that family, what is the chance the child is female?

$(d)$ Given that the child chosen from the sampling scheme of part $(c)$ is female, what is the chance the child is from a family with 2 children?

$(e)$ Given that the child chosen from the sampling scheme of part $(c)$ is female, what is the chance she has an older sister?

My initial approach to part (c) was as follows:

$\frac{1}{4} \times\ \frac{7}{11} = \frac{7}{44}$

but this seems too simple and I feel like there is more to it that I'm not seeing...

As for part (e), I understand that only $3$ of the females in the sample space could have older sisters, but I am stuck on the calculations for this one.

• I was a pretty random child, by any measure ... (but not female). – wolfies Mar 13 '16 at 13:26

A certain country town has 4 families having respectively 4,3,2,2 children with genders FFMM, FFM, FF, FM.

(c) If a random family is chosen from the town and then a random child from that family, what is the chance the child is female?

My initial approach to part (c) was as follows:

$$\tfrac 14\times \tfrac 7{11}=\tfrac 7{44}$$

but this seems too simple and I feel like there is more to it that I'm not seeing...

Close, but yes, you are missing something.   The probability of choosing the first family, and a girl from that family is $\tfrac 1 4\cdot \tfrac 2 4$, and similarly for the other three families.

$$\mathsf P(\textsf{female})~=~\frac 14(\frac 24+\frac 23+\frac 22+\frac 12)~=~ \frac 2 3$$

You can't just compare the ratio of female to all-children because there is not an equal chance of selecting each child.

For part $d$ and $e$ use Bayes' rule.

$$\mathsf P(\textsf{2-child-family}\mid \textsf{female}) ~=~\frac{\mathsf P(\textsf{2-child-family}\cap \textsf{female})}{\mathsf P(\textsf{female})}$$

$$\mathsf P(\textsf{has-elder-sister}\mid \textsf{female})~ =~\frac{\mathsf P(\textsf{has-elder-sister}\cap \textsf{female})}{\mathsf P(\textsf{female})}$$

• I see, it makes sense now! Using the results from part (c), I found: > $P(family-of-2|F) = [ \frac{1}{4} + \frac{1}{2} ] / \frac{2}{3}$ $= \frac{9}{16}$ Would this be correct? – Stephf97 Mar 13 '16 at 12:44
• and similarly I found $\frac{13}{32}$ for part (e) – Stephf97 Mar 13 '16 at 12:49
• Yes, assuming that was a typo as it is: $\frac{\frac 1 4(1+\frac 1 2)}{\tfrac 2 3}=\frac 9{16}$; also yes for part e – Graham Kemp Mar 13 '16 at 12:51