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In a particular town 10% of the families have no children, 20% have one child, 40% have two children, 20% have three children, and 10% have four. Let T represent the total number of children, and G the number of girls, in a family chosen at random from this town. Assuming that children are equally likely to be boys or girls, find the distribution of G. Please Display your answer in a table.

I tried, this but my set up didn't help me: P(G = n| G = B = 1/2) where n is the number of girls, and probability of G(Girl) = B(boy) = 1/2. I used conditional probabilities, but didn't help me, setup? Also, I can say that the distribution when G = 0 is 0. How to incorporate the number of childrens total? I only used the fact that we are interested in the number of girls, and boy and girl is equally likely

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$G=B=1/2$ means nothing, since $G$ and $B$ are random integers. Hint 1: What are $P(T=n)$ and $P(G=0|T=n)$, for each $0\le n\le 4$. Hint 2: Deduce the value of $P(G=0)$ from the answer to hint 1 and from Bayes' total probability formula en.wikipedia.org/wiki/Bayes%27_theorem. Then replace $0$ by any $k$ between $0$ and $4$ and repeat. –  Did Apr 17 '11 at 10:33
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Homework questions should be tagged with the "homework" tag. –  Michael Lugo Apr 18 '11 at 3:26

1 Answer 1

up vote 3 down vote accepted

I will begin the calculation, and you can complete it. The random variable $G$ (the number of girls) can take on no values other than $0$, $1$, $2$, $3$, and $4$. The easiest thing to calculate is $P(G=4)$.

How can there be $4$ girls? We must be dealing with a family of $4$ children (probability $0.1$) and that family must have $4$ girls. The probability of $4$ girls given that there are $4$ children is $1/2^4$. So the probability that a family has $4$ children and they are all girls is $(0.1)(1/16)$. To prepare for later, let's do this again with symbols. $$P((T=4)\cap (G=4))=P((G=4)|(T=4))P(T=4)=(1/2^4)(0.1)$$ Now let us calculate the probability that $G=3$. This event can happen in two ways: (a) The family has $3$ children and they are all girls OR (b) the family has $4$ children and exactly $3$ of them are girls.

First tackle (a). Let us find the probability of $3$ children, all girls. The probability of $3$ children is $0.2$. Given that a family has $3$ children, the probability they are all girls is $1/8$. So the probability of $3$ children and all girls is $(0.2)(1/8)$. Now tackle (b). The probability of $4$ children is $(0.1)$. Given that there are $4$ children, the probability there are exactly $3$ girls is $4/16$ (I hope this part is known to you). So the probability of $4$ children and exactly $3$ girls is $(0.1)(4/16)$. Add up the two probabilities we obtained. The probability of $3$ girls is $(0.2)(1/8)+(0.1)(4/16)$. In symbols, $$P(G=3)= P((G=3)|(T=3))P(T=3)+P((G=3)|(T=4))P(T=4)$$

Now you can proceed to find the probability of $2$ girls ($3$ cases) also $1$ girl, also $0$ girls. If you have calculated $4$ probabilities, the fifth can be found by subtracting the sum of the others from $1$. But you might as well calculate all $5$ probabilities, and use the fact they must add up to $1$ as a check of your calculations.

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For 3 children, I think you mean: 0.1(1/4) + 0.2(1/8)? –  mary Apr 17 '11 at 19:39
    
@user8917: Thank you, a slip, now fixed. And I used the $4/16$ instead of the suggested $1/4$ because the $4/16$ keeps more of the structure. –  André Nicolas Apr 17 '11 at 21:41

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