Probability- Brain Teaser- 1 boy and 1 girl

In summary, the conversation was about finding the expected number of children in a family that stops having children once they have at least one boy and one girl. The approach to finding the expected value involved using a sum and taking the derivative of a geometric sum. After correcting a mistake, the final answer was determined to be 3.
  • #1
Roni1985
201
0

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
[tex]\sum n(\frac{1}{2})^n [/tex]
n=2
 
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  • #2
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.
 
  • #3
Dick said:
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
[tex]\sum n(1/2)^(^n^-^1^)[/tex]
n=2

Now, I know the sum of the derivative of a geometric sum is
[tex]\frac{a}{(1-r)^2}[/tex]
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
[tex]\sum r^n[/tex]= [tex]\frac{r^2}{1-r}[/tex]
n=2

I need to take the derivative of this and I get [tex]\frac{2r-r^2}{(1-r)^2}[/tex]
I plug in 1/2 and multiply the answer by 1/2 that I pulled out of the sum before.
 
Last edited:
  • #4
still looking for help :)

Thanks.
 
  • #5
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?
 
Last edited:
  • #6
Dick said:
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
[tex]\sum (n+1)(1/2)^n[/tex]
n=1

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

inf
[tex]\sum \frac{d}{dr}(r)^(^n^+^1^)[/tex]
n=1

inf
[tex]\frac{d}{dr}[/tex][tex]\sum(r)^(^n^+^1^)[/tex]
n=1

inf
[tex]\frac{d}{dr}[/tex]*r[tex]\sum(r)^n[/tex] = [tex]\frac{d}{dr}[/tex](*r*[tex]\frac{r}{1-r}[/tex])=[tex]\frac{2r-r^2}{(1-r)^2}[/tex]
n=1

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

Am I doing something wrong?
 
  • #7
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.
 
  • #8
Roni1985 said:
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.
 
  • #9
Dick said:
Now it's fine. I was just wondering how you got 3/2.

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

1) What is the probability of having a boy and a girl in a family with two children?

The probability of having a boy and a girl in a family with two children is 50%, or 1 in 2. This is because each child has an equal chance of being born as either a boy or a girl, and the outcomes are independent of each other.

2) If a family has two children and one is a boy, what is the probability that the other child is also a boy?

The probability that the other child is also a boy is 50%, or 1 in 2. This may seem counterintuitive, but it is important to remember that the outcomes of each child's gender are independent, regardless of what the other child's gender is.

3) What is the probability of having two boys in a family with two children?

The probability of having two boys in a family with two children is 25%, or 1 in 4. This is because there are four possible outcomes when flipping a coin twice (BB, BG, GB, GG) and only one of those outcomes has two boys.

4) In a group of four children, what is the probability of having two boys and two girls?

The probability of having two boys and two girls in a group of four children is 37.5%, or 3 in 8. This can be calculated by using the binomial distribution formula, where n=4, k=2, and p=0.5 (for equal chance of boy or girl).

5) Is the probability of having a boy and a girl in a family with two children affected by the order of their births?

No, the probability is not affected by the order of their births. Each child's gender is determined independently of their sibling's gender, so the probability remains the same regardless of which child is born first.

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