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)
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.
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
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 =
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.
We had:
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)
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. ??