0
$\begingroup$

I have the following problem,

Suppose that 15 % of the families in a certain community have no children, 20 % have one, 35 % have two and 30 % have three. Suppose further, that in each family each child independently is equally likely to be a boy or a girl. Assume a family in this community is chosen at random and let X be the number of girls and Y be the number of boys in that family.

I try to understand the solution given,

Table:enter image description here

And the process said:

P{B = 0, G = 0} = P{no children} = 0.15;


P{B = 0, G = 1} = P{1 girl and a total of 1 child}

= P{1 child}×P{1 girl|1 child}

= 0.20×0.50 = 0.1;

I belive, I understand this because the possibles outcomes are (0,1)(1,0) only 1 of 2 =1/2


P{B = 0, G = 2} = P{2 girl and a total of 2 children} = P{2 children}×P{2 girls|2 children} = 0.35×(0.50)^2 = 0.0875,

I can't see this, because my tought is Total outcomes in 2 children (1,1)(2,0) and (0,2) therefor 1/3 of the 0.35 probability.

I know that something is wrong in my thought but I can't see it. If someone explains the correct way, I will be tanks.

$\endgroup$
1
$\begingroup$

The outcomes with two children are actually BB, BG, GB, and GG, where G stands for a girl, B for a boy, the first symbol gives the sex of the first child, and the second symbol gives the sex of the second child. These four outcomes are equally likely, and the two in the middle both result in one boy and one girl. Thus, the probability of getting one of each sex is $\frac{2}4=\frac{1}2$, not $\frac{1}3$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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