Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is my probabilistic for the problem of determining the probability, that a arbitrary chosen boy has a sister, if it is equal likely that a family has boys or girl and if the probability of having $0,1,\ldots,4$ children are $p_0,\ldots,p_4$ correct?

My idea was that $\Omega= \{n,\ b,g,\ bb,gg,bg,\ bbb,bbg,bgg,ggg,\ldots \}$ (where $n$ denotes the fact that the family has no children, $b$ stands for "boy" and $g$ for "girl") and $P(n)=p_0,P(bb,gg,bg,)=p_1$ and so on - and since boys and girls are equal like we get $P(bb)=P(gg)=P(bg)=\frac{p_1}{3}$ and so on.

share|improve this question
    
Not much detail has been given. The sample space would probably work. Already some of the probabilities you have given are wrong. In a two-child family, under usual assumptions, the probability of two boys is $1/4$, as is the probability of two girls. The probability of mixed is $1/2$. –  André Nicolas Mar 14 '12 at 22:32

2 Answers 2

We will make the usual unrealistic independence assumptions. The stated assumptions of the problem are not enough. For example, we can perfectly well have equal probability of boy and girl on any birth, but a family stops having children once a boy is born. That would not affect overall sex distribution, but would have a strong effect on the makeup of families.

Imagine choosing a child at random. To make formulas simpler, let $$T=p_1+2p_2+3p_3+4p_4.$$ Then the probability the child is an only child is $\frac{p_1}{T}$, the probability the child comes from a $2$-child family is $\frac{2p_2}{T}$, the probability the child is from a $3$-child family is $\frac{3p_3}{T}$, and the probability the child is from a $4$-child family is $\frac{4p_4}{T}$.

By symmetry, the probability that a randomly chosen boy has a sister is the same as the probability that a randomly chosen child has a sibling of the opposite sex. Things will be easier to visualize if we imagine that the chosen child is the oldest (it makes no difference).

We first calculate the probability that the chosen child does not have a sibling of the opposite sex. List the sequence of sexes of children, oldest first.

If the child is from a $1$-child family, the probability of having no sibling of the opposite sex is $1$.

If the child is from a $2$-child family, things get more complicated. There are $2$ possible sex orders once we know the sex of the oldest. For $1$ of these, there is no sibling of the opposite sex. So the probability the oldest has no sibling of the opposite sex is $1/2$.

If the oldest child is from a $3$-child family, there are $4$ equally likely sex distributions with the sex of the oldest child given. The oldest has no sibling of the opposite sex in only $1$ of these cases. So our probability is $1/4$.

Finally, there is a probability of only $1/8$ that the oldest child in a $4$-child family has no sibling of the opposite sex.

Add up. The probability that the oldest has no sibling of the opposite sex is $$\frac{p_1}{T}+ \frac{1}{2}\cdot\frac{2p_2}{T}+\frac{1}{4}\cdot\frac{3p_3}{T}+\frac{1}{8}\cdot\frac{4p_4}{T}.$$

Finally, the probability that a randomly chosen child, say a boy, though it doesn't matter has a sibling of the opposite sex is obtained by subtracting the expression in $(\ast)$ from $1$.

share|improve this answer

Hint: Break it by cases. Given that the family has one child: Probability of a boy having a sister is zero.

Given that the family has 2 children, then probability that a boy picked from that family has a sister is $\frac{1}{2}$, and more generally:

$$P(\mbox{boy having a sister})=\sum_{i}P(\mbox{boy having a sister}|\mbox{the family has }i \mbox{ kids})p_i$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.