Probability- Brain Teaser- 1 boy and 1 girl

[Roni1985]

Homework Statement

I'm not sure if this is the exact wording, but here it is:
Let's assume a family wants at least 1 girl and 1 boy. It stops having kids once it get both.

Ex:
BBBBBG
GB
GGGGGGGGGB
etc.

What is the expect number of kids in the family?

Homework Equations

Is this a geometric random variable?

The Attempt at a Solution

I attempted to find the expected value by using a sum.
I can find the sum of the probabilities and I'm getting 1, but can't seem to find the expected value.
inf
$$\sum n(\frac{1}{2})^n$$
n=2

 

[Dick]

The probability of stopping after 2 children is 1/2, isn't it? That doesn't fit very well with your (1/2)^n. You could have BG, GB or BB, GG. Both are equally likely. It's not a strictly geometric sum. But it's closely related. It's the derivative of a geometric sum.

 

[Roni1985]

The probability of stopping after 2 children is 1/2, isn't it? That doesn't fit very well with your (1/2)^n. You could have BG, GB or BB, GG. Both are equally likely. It's not a strictly geometric sum. But it's closely related. It's the derivative of a geometric sum.

Hello,

I found my mistake, this is what I'm getting
inf
$$\sum n(1/2)^(^n^-^1^)$$
n=2

Now, I know the sum of the derivative of a geometric sum is
$$\frac{a}{(1-r)^2}$$
However, my sum starts from n=2 and by doing the calculations, I think I'm getting 3/2.
Can you please confirm my answer?
inf
$$\sum r^n$$= $$\frac{r^2}{1-r}$$
n=2

I need to take the derivative of this and I get $$\frac{2r-r^2}{(1-r)^2}$$
I plug in 1/2 and multiply the answer by 1/2 that I pulled out of the sum before.

 

[Roni1985]

still looking for help :)

Thanks.

 

[Dick]

Let's do a sanity check. Your sum is 2*(1/2)+3*(1/4)+4*(1/8)+5*(1/16)+6*(1/32)+... Evaluate that. You've got the series right. Your sum is wrong. Just sum the first two terms. That's already over 3/2. I don't really understand how you got 3/2. Can you explain that again? Where did you pull out a '1/2'? Differentiating r^2/(1-r) is on the right track. How did you get that?

 

[Roni1985]

Let's do a sanity check. Your sum is 2*(1/2)+3*(1/4)+4*(1/8)+5*(1/16)+6*(1/32)+... Evaluate that. You've got the series right. Your sum is wrong. Just sum the first two terms. That's already over 3/2. I don't really understand how you got 3/2. Can you explain that again? Where did you pull out a '1/2'? Differentiating r^2/(1-r) is on the right track. How did you get that?

I don't seem to get it.
What do you mean my sum is wrong?

I can start from 1.
inf
$$\sum (n+1)(1/2)^n$$
n=1

This is also the derivative of a geometric sum.
Now,

inf
$$\sum \frac{d}{dr}(r)^(^n^+^1^)$$
n=1

inf
$$\frac{d}{dr}$$$$\sum(r)^(^n^+^1^)$$
n=1

inf
$$\frac{d}{dr}$$*r$$\sum(r)^n$$ = $$\frac{d}{dr}$$(*r*$$\frac{r}{1-r}$$)=$$\frac{2r-r^2}{(1-r)^2}$$
n=1

Now, I can plug in 1/2 and actually now I'm getting 3.

Am I doing something wrong?

 

[Roni1985]

Oh, just found a TI-89 emulator for Windows, so I calculated the sum and it is 3.
Can you just look at my way. Is everything correct ?
Thanks.

 

[Dick]

Oh, just found a TI-89 emulator for Windows, so I calculated the sum and it is 3.
Can you just look at my way. Is everything correct ?
Thanks.

Now it's fine. I was just wondering how you got 3/2.

 

[Roni1985]

Now it's fine. I was just wondering how you got 3/2.

Yes, I made a mistake before.
Thanks for your help.