# Randomly chosen family

• Oct 7th 2011, 11:29 PM
stevetall
Randomly chosen family
A number of children in the families is 0 - 4. Children and probability are:
0 child = 0.15,
1 = 0.25,
2 = 0.3,
3 = 0.2,
4 = 0.1
Question: What is the probability of this randomly chosen family having exactly 2 girls?

I think we must consider total probability.
- And there are 8 ways that out of four children 1 are girls 4!/(4!*1!)
- 8 ways that out of four children 2 are girls 4!/(3!*2!)
- 8 ways that out of four children 3 are girls 4!(2!*3!)
- 5 ways that out of four children 4 are girls 4!(1!*4!)
Boys and girls do not correlate, they are independent. Probability that there is a girl is 1/2. We choose randomly one family ( which could have 0 or 1 or 2 or 3 or 4 children ).

So two girls in a randomly chosen family has over 30? possibilities
And of these 0-4 children having families are exactly 2 girls
probability is 4/?
Unless this is 5/? - because 0,1,2,3,4 = 5 different alternatives.

Some thoughts;
I think this is not needed: 0.15 * 0.25 * 0.3 * 0.2 * 0.1 ...
BUT: this could be what is needed here: P(B) = P(A) P(B|A) + P(A^c) P(B|A^c)
• Oct 7th 2011, 11:48 PM
CaptainBlack
Re: Randomly chosen family
Quote:

Originally Posted by stevetall
A number of children in the families is 0 - 4. Children and probability are:
0 child = 0.15,
1 = 0.25,
2 = 0.3,
3 = 0.2,
4 = 0.1
Question: What is the probability of this randomly chosen family having exactly 2 girls?

I think we must consider total probability.
- And there are 8 ways that out of four children 1 are girls 4!/(4!*1!)
- 8 ways that out of four children 2 are girls 4!/(3!*2!)
- 8 ways that out of four children 3 are girls 4!(2!*3!)
- 5 ways that out of four children 4 are girls 4!(1!*4!)
Boys and girls do not correlate, they are independent. Probability that there is a girl is 1/2. We choose randomly one family ( which could have 0 or 1 or 2 or 3 or 4 children ).

So two girls in a randomly chosen family has over 30? possibilities
And of these 0-4 children having families are exactly 2 girls
probability is 4/?
Unless this is 5/? - because 0,1,2,3,4 = 5 different alternatives.

Some thoughts;
I think this is not needed: 0.15 * 0.25 * 0.3 * 0.2 * 0.1 ...
BUT: this could be what is needed here: P(B) = P(A) P(B|A) + P(A^c) P(B|A^c)

P(exactly 2 girls)=P(exactly two girls | 2 children)P(2 children)+P(exactly two girls | 3 children)P(3 children)+P(exactly two girls | 4 children)P(4 children)

The conditional probabilities can be evaluated using the appropriate binomoal distributions.

CB
• Oct 9th 2011, 07:55 AM
stevetall
Re: Randomly chosen family
Indeed, the conditional probabilities are the right tools here!

Here is my new effort.

P(A) = exactly 2 girls P(A^c)= 2 children P(B) = 3 children P(C) = 4 children

P(A) = P(A | A^c ) P(A^c ) + P(A | B) P(B ) + P(A | C) P(C)

And probabilitie were; ( so 0 child = 0.15 and 1 = 0.25 are not needed in this )

2 children = 0.3,
3 children = 0.2,
4 children = 0.1

So could it be:
P(A) = P(A | 0.3 ) P(0.3 ) + P(A | 0.2) P(0.2 ) + P(A | 0.1) P(0.1)

But I don't know how a solution should be done.
• Oct 9th 2011, 09:51 AM
CaptainBlack
Re: Randomly chosen family
Quote:

Originally Posted by stevetall
Indeed, the conditional probabilities are the right tools here!

Here is my new effort.

P(A) = exactly 2 girls P(A^c)= 2 children P(B) = 3 children P(C) = 4 children

P(A) = P(A | A^c ) P(A^c ) + P(A | B) P(B ) + P(A | C) P(C)

And probabilitie were; ( so 0 child = 0.15 and 1 = 0.25 are not needed in this )

2 children = 0.3,
3 children = 0.2,
4 children = 0.1

So could it be:
P(A) = P(A | 0.3 ) P(0.3 ) + P(A | 0.2) P(0.2 ) + P(A | 0.1) P(0.1)

But I don't know how a solution should be done.

Well you should be able to compute the first of these; the probability that if you have two children both are girls.

But all of them are subject to the same method:

P(N girls in M children)=b(N;M,0.5)

where b(.;.,.) is the binomial distribution mass function.

CB
• Oct 9th 2011, 09:47 PM
stevetall
Re: Randomly chosen family
So it was; P(A) = exactly 2 girls P(A^c)= 2 children P(B) = 3 children P(C) = 4 children

P(A) = P(A | A^c ) P(A^c ) + P(A | B) P(B ) + P(A | C) P(C)

and here's new.

probability that in two children both are girls = 1/2 * 1/2 = 1/4

probability that in three children two are girls = 1/2 * 1/2 * 1/2 = 1/6

probability that in four children two are girls = 1/2 * 1/2 * 1/2 * 1/2 = 1/8
• Oct 10th 2011, 12:03 AM
CaptainBlack
Re: Randomly chosen family
Quote:

Originally Posted by stevetall
So it was; P(A) = exactly 2 girls P(A^c)= 2 children P(B) = 3 children P(C) = 4 children

P(A) = P(A | A^c ) P(A^c ) + P(A | B) P(B ) + P(A | C) P(C)

and here's new.

probability that in two children both are girls = 1/2 * 1/2 = 1/4

probability that in three children two are girls = 1/2 * 1/2 * 1/2 = 1/6

probability that in four children two are girls = 1/2 * 1/2 * 1/2 * 1/2 = 1/8

>>binomial distribution<<

CB
• Oct 10th 2011, 12:21 AM
stevetall
Re: Randomly chosen family
I read it and try once again.

So it was; P(A) = exactly 2 girls P(A^c)= 2 children P(B) = 3 children P(C) = 4 children

P(A) = P(A | A^c ) P(A^c ) + P(A | B) P(B ) + P(A | C) P(C)

and here's new.

probability that in two children both are girls = 1/4 * 1/4 = 1/8

probability that in three children two are girls = 1/4 * 1/4 * 1/4 =

probability that in four children two are girls = 1/4 * 1/4 * 1/4 * 1/4 =
• Oct 10th 2011, 10:59 PM
stevetall
Re: Randomly chosen family
I found errors and I had forgot to put rest of the results. I am terribly sorry, but I have really tried to solve this.

P(A) = exactly 2 girls P(A^c)= 2 children P(B) = 3 children P(C) = 4 children

P(A) = P(A | A^c ) P(A^c ) + P(A | B) P(B ) + P(A | C) P(C)

And my solution is:

probability that in two children both are girls = 1/4 + 1/4 = 1/8

probability that in three children two are girls = 1/4 + 1/4 + 1/4 = 1/12

probability that in four children two are girls = 1/4 + 1/4 + 1/4 + 1/4 = 1/16

AND: probability that there are exactly 2 girls =

P(A) = exactly 2 girls P(A^c)= 2 children P(B) = 3 children P(C) = 4 children

P(A) = P(A | A^c ) P(A^c ) + P(A | B) P(B ) + P(A | C) P(C)

I am not sure how to use this: P(N girls in M children)=b(N;M,0.5)
• Oct 10th 2011, 11:53 PM
CaptainBlack
Re: Randomly chosen family
Quote:

Originally Posted by stevetall
I found errors and I had forgot to put rest of the results. I am terribly sorry, but I have really tried to solve this.

P(A) = exactly 2 girls P(A^c)= 2 children P(B) = 3 children P(C) = 4 children

P(A) = P(A | A^c ) P(A^c ) + P(A | B) P(B ) + P(A | C) P(C)

And my solution is:

probability that in two children both are girls = 1/4 + 1/4 = 1/8

probability that in three children two are girls = 1/4 + 1/4 + 1/4 = 1/12

probability that in four children two are girls = 1/4 + 1/4 + 1/4 + 1/4 = 1/16

AND: probability that there are exactly 2 girls =

P(A) = exactly 2 girls P(A^c)= 2 children P(B) = 3 children P(C) = 4 children

P(A) = P(A | A^c ) P(A^c ) + P(A | B) P(B ) + P(A | C) P(C)

I am not sure how to use this: P(N girls in M children)=b(N;M,0.5)

The part in red is nonsense the part in blue is how to compute these.

CB
• Oct 11th 2011, 12:34 AM
stevetall
Re: Randomly chosen family
b(N;M,0.5) - i will try.

( N over 0.5 ) =

2 children:

2!
____

0.5!(2-0.5)!

= 2/1 = 2

3 children:

3!
____

0.5!(3-0.5)!

= 6/2 = 3

4 children:

4!
____

0.5!(4-0.5)!

= 24/3 = 8
• Oct 11th 2011, 09:23 PM
stevetall
Re: Randomly chosen family
I give up. I don't know what a solution is and how that must be done, too difficult.
• Oct 11th 2011, 10:07 PM
stevetall
Re: Randomly chosen family
But I want to tell you what I achieved.

Let's mark a girl with g and a boy with b. Because there could be four children, then there have to be
2 * 2 * 2 * 2 = 16

gggg
gggb
ggbb
ggbg

bbbb
bbbg
bbgg
bbgb

bggg
bggb
bgbg
bgbb

gbgg
gbgb
gbbg
gbbb

I find 7 hits, so probability that there are two girls is 7/16. ??
• Oct 12th 2011, 05:22 AM
CaptainBlack
Re: Randomly chosen family
Quote:

Originally Posted by stevetall
b(N;M,0.5) - i will try.

( N over 0.5 ) =

2 children:

2!
____

0.5!(2-0.5)!

= 2/1 = 2

3 children:

3!
____

0.5!(3-0.5)!

= 6/2 = 3

4 children:

4!
____

0.5!(4-0.5)!

= 24/3 = 8

$b(N;M,p)={M \choose N} p^n(1-p)^{M-N}=\frac{M!}{N!(M-N)!}p^n(1-p)^{M-N}$

Alternative notation:

$b(N;M,p)={_MC_N}\; p^n(1-p)^{M-N}}=\frac{M!}{N!(M-N)!}p^n(1-p)^{M-N}$

$b(2;2,0.5)=\frac{2!}{2!0!}(0.5)^2=0.25$

$b(2;3,0.5)=\frac{3!}{2!(3-2)!}(0.5)^2(0.5)^1=3 \times 0.5^3=0.375$

$b(2;4,0.5)=\frac{4!}{2!(4-2)!}(0.5)^2(0.5)^2=2 \times 3 \times 0.5^4=0.375$

See the first shaded block at the link given, also scroll down for a binomial calculator.

CB
• Oct 12th 2011, 10:06 AM
stevetall
Re: Randomly chosen family
That was fantastic!, really great work. Thank you! Really interesting solution, it sure was too hard for me. But I really tried.

And I visited that link again interesting - now I tried it myself.