# Thread: In a family of 5 children, determine the following probabibilities?

1. ## In a family of 5 children, determine the following probabibilities?

Hi, I'm a bit confused with one part of this probability problem. I'm pretty sure I got part a and b, but I'm not sure about c.

A family has 5 children, determine the following probabilities

a.) All 5 are the same gender

b.) 3 are boys, and 2 are girls

c.) The 3 oldest are boys, and the 2 youngest are girls.

For a. I'm pretty sure it's 1/32 since it would be .5^5 b/c there's a 1/2 probability of being either gender for each of the 5 children.
For b. I think it's also 1/32 since the probability of either gender doesn't change per birth.
As far as c. is concerned I think it's 5/16 because C(5,3) = 10 ways to have 3 boys and 2 girls multiplied by 1/32 but I'm not sure

2. ## Re: In a family of 5 children, determine the following probabibilities?

Originally Posted by crownvicman
Hi, I'm a bit confused with one part of this probability problem. I'm pretty sure I got part a and b, but I'm not sure about c.

A family has 5 children, determine the following probabilities

a.) All 5 are the same gender

b.) 3 are boys, and 2 are girls

c.) The 3 oldest are boys, and the 2 youngest are girls.

For a. I'm pretty sure it's 1/32 since it would be .5^5 b/c there's a 1/2 probability of being either gender for each of the 5 children.
For b. I think it's also 1/32 since the probability of either gender doesn't change per birth.
As far as c. is concerned I think it's 5/16 because C(5,3) = 10 ways to have 3 boys and 2 girls multiplied by 1/32 but I'm not sure
a) is correct

b) you neglect to count the various ways you can have 3 boys and 2 girls, think of a 5 digit binary string. There are more than one with 3 1's and 2 0's

c) did you mix up b and c? In this case there is only one string, if you order it by age, that matches the criterion, 11100, out of the 32 possible strings. So in this case the probability is just

$\left(\frac{1}{2}\right)^5=\frac{1}{32}$

If you think of this as a probability tree you are being forced down a single path through the tree in contrast with b.

3. ## Re: In a family of 5 children, determine the following probabibilities?

Thank you, sorry about that I did flip flop what I meant to say for b and c. Is 5/16 the correct answer for 3 are boys and 2 are girls?

4. ## Re: In a family of 5 children, determine the following probabibilities?

yes 5/16 is right

5. ## Re: In a family of 5 children, determine the following probabibilities?

Originally Posted by crownvicman
A family has 5 children, determine the following probabilities
a.) All 5 are the same gender
To model this think of a truth table with five inputs.
Replace the T's with g's & the F's with b's. Now that is a model of the possible the birth-order and gender of five children. There are thirty-two rows, the first is all g's and the last is all b's.

Thus, there are two out of thirty-two of $\frac{1}{16}$ probability of same gender.

6. ## Re: In a family of 5 children, determine the following probabibilities?

So does this mean the answer to part a.) is actually 1/16 since the probability is 1/32 for only 1 gender yet there are two gender possibilities so that 1/32 is multiplied by 2?

then b.) is still 5/16

and c.) is still 1/32 for 3 boys, 2 girls in any order?

7. ## Re: In a family of 5 children, determine the following probabibilities?

Originally Posted by crownvicman
So does this mean the answer to part a.) is actually 1/16 since the probability is 1/32 for only 1 gender yet there are two gender possibilities so that 1/32 is multiplied by 2?

then b.) is still 5/16

and c.) is still 1/32 for 3 boys, 2 girls in any order?
man I really screwed the pooch on this one.

a) is as Hollywood noted 2 x 1/32 = 1/16

b) is 5/16

for (c) there are 3!=6 ways to arrange the 3 boys as the 3 oldest, and then 2 ways to arrange the girls in the remaining 2 slots. So there are 12 ways altogether to arrange the 3 boys and 2 girls such that the 3 boys are the oldest.

So the overall probability of this configuration is 12/32 = 3/8

8. ## Re: In a family of 5 children, determine the following probabibilities?

Originally Posted by romsek
(c) there are 3!=6 ways to arrange the 3 boys as the 3 oldest, and then 2 ways to arrange the girls in the remaining 2 slots. So there are 12 ways altogether to arrange the 3 boys and 2 girls such that the 3 boys are the oldest.

So the overall probability of this configuration is 12/32 = 3/8
There is only one way to have three b's followed by two g's.
So $\dfrac{1}{32}~.$

9. ## Re: In a family of 5 children, determine the following probabibilities?

that's not true. we're not given any rule for assigning ages so we can order them however we like.

let the boy's be b1 b2 b3, the girls g1 and g2

b1 b2 b3 g1 g2

and

b3 b2 b1 g1 g2

for example both fit the bill

10. ## Re: In a family of 5 children, determine the following probabibilities?

Originally Posted by romsek
that's not true. we're not given any rule for assigning ages so we can order them however we like.
What I posted is correct

Suppose we have blue balls and green balls that are identical except for color.
Say ten of each color, or maybe a hundred.

We select five of them and randomly arrange them into a string.
What is the probability that the string turns out to be $BBBGG~?$

That is the model for this question. It is like flipping a coin five times.
$\mathcal{P}(HHHTT)=~?$

11. ## Re: In a family of 5 children, determine the following probabibilities?

Originally Posted by Plato
What I posted is correct

Suppose we have blue balls and green balls that are identical except for color.
Say ten of each color, or maybe a hundred.

We select five of them and randomly arrange them into a string.
What is the probability that the string turns out to be $BBBGG~?$

That is the model for this question. It is like flipping a coin five times.
$\mathcal{P}(HHHTT)=~?$
balls are indistinguishable, children are

12. ## Re: In a family of 5 children, determine the following probabibilities?

Originally Posted by romsek
balls are indistinguishable, children are
Balls are indistinguishable, children are either boy or girl.

13. ## Re: In a family of 5 children, determine the following probabibilities?

Originally Posted by Plato
Balls are indistinguishable, children are either boy or girl.
Well I can't argue with you over your interpretation of the question. If you say the children are indistinguishable then everything you've said is correct.

I'll just say that it's pretty obvious to me that children are more than just their gender and that if the problem wanted to make the objects indistinguishable it should have used balls or the like.

14. ## Re: In a family of 5 children, determine the following probabibilities?

Originally Posted by romsek
Well I can't argue with you over your interpretation of the question. If you say the children are indistinguishable then everything you've said is correct.
Originally Posted by crownvicman
A family has 5 children, determine the following probabilitie
a.) All 5 are the same gender
b.) 3 are boys, and 2 are girls
c.) The 3 oldest are boys, and the 2 youngest are girls.
That question is not about the children. It is about the gender of the five siblings.
It is not about individual children.

15. ## Re: In a family of 5 children, determine the following probabibilities?

Originally Posted by Plato
That question is not about the children. It is about the gender of the five siblings.
It is not about individual children.
First off let's make sure you understand I"m not trying to argue with you for the sake of arguing with you. I'd like to understand this.

For (a) and (b) I agree with you. But (c) now implies an ordering on the children by age. Thus they do have at least the further property of age. If I can order them by age I can individually tag them, unless two happen to be born at precisely the same time in which case the ordering isn't well defined.

If I can individually tag them then each of them is distinguishable via their tag.

What am I getting wrong?

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