Ms. Barton has four children. You are told correctly that she has at

Hi people,

Suddenly, something comes to my mind so awfully, but I couldn't disprove why I am wrong. Could someone help me in this.

I suddenly see it as : P(2 girls, 2 boys) = 1/2*1/2*1/2*1/2 = 1/16.(since each has a prob of 1/2).

Pls disprove me.... I don see where I am wrong, but I know surely I am.

Thanks

What am I missing here?

The questions clearly states that 2 of the 4 kids are girls. It also clearly asks "what is the probability that the other 2 are boys?"

This gives us a P of 1/2 that a child will be a boy and 1/2 that the child will be a girl. Armed with the fact that having two girls will have no significance on the sex of the other two children, we can treat this as mutually exclusive scenario.

So there is P:1/2 that one of the remaining children is a boy, and P:1/2 that the other child is also a boy. Therefore, the probability would become 1/4.

Am I reading this question wrong?

The following is the complete explanation for this question:

You need to find Probability of 2 boys and 2 girls given there are at least 2 girls. This is a conditional probability. You have already been given that there are at least 2 girls. You need to find the probability of 2 boys and 2 girls given this condition.
P(A given B) = P(A)/P(B)

P(2 boys, 2 girls) = (1/2)*(1/2)*(1/2)*(1/2) * 4!/(2!*2!) = 3/8
You multiply by 4!/(2!*2!) because out of the four children, any 2 could be boys and the other two would be girls. So you have to account for all arrangements: BGBG, BBGG, BGGB etc

P(atleast 2 girls) = P(2 boys, 2 girls) + P(1 boy, 3 girls) + P(4 girls) = 1 - P(4 boys) - P(3 boys, 1 girl)
You can solve P(atleast 2 girls) in any way you like.
P(1 boy, 3 girls) = (1/2)*(1/2)*(1/2)*(1/2) * 4!/3! = 1/4
P(4 girls) = (1/2)*(1/2)*(1/2)*(1/2) = 1/16
P(atleast 2 girls) = 3/8 + 1/4 + 1/16 = 11/16

P(2 boys, 2 girls given atleast 2 girls) = P(2 boys, 2 girls) / P(atleast 2 girls) = (3/8)/(11/16) = 6/11

In the above given solutions, all the (1/2)s have been ignored because the probability of a boy is 1/2 and of a girl is 1/2 too. Hence in every case you will get (1/2)*(1/2)*(1/2)*(1/2) which can be ignored. We just need to focus on arrangements in each case. This is similar to the famous coin flipping questions (the probability of 3 Heads and a tail etc) (check out Binomial Probability in Veritas Combinatorics book for an explanation)

Hi Karishma,
Perhaps you missed my previous question.
I don't understand why when calculating the odds of calculating 2 boys and 2 girls, you multiply by (1/2)^4. We are already told that 2 girls are given... So why not multiply just by the odds of the rest being boys, which is (1/2)^2?

Whats the significance of the following statement in the question- "Assume that the probability of having a boy is the same as the probability of having a girl." ?

Ms. Barton has four children. You are told correctly that she has at least two girls but you are not told which two of her four children are those girls. What is the probability that she also has two boys? (Assume that the probability of having a boy is the same as the probability of having a girl.)

(A) 1/4

(B) 3/8

(C) 5/11

(D) 1/2

(E) 6/11


We are concerned about the 4 selections together, not the individual selection. What this means is that if your logic is correct, the probability of having 4 girls would be the same as the probability of having 2 girls and 2 boys. But is that correct? No it is not.

In how many ways can you have 4 girls?
(1/2)*(1/2)*(1/2)*(1/2) = 1/16 (First child girl,next girl, next girl, next girl)

In how many ways can you have 2 girls and 2 boys?
(1/2)*(1/2)*(1/2)*(1/2) = 1/16 (First child girl,next girl, next boy, next boy)
(1/2)*(1/2)*(1/2)*(1/2) = 1/16 (First child girl,next boy, next girl, next boy)
(1/2)*(1/2)*(1/2)*(1/2) = 1/16 (First child girl,next boy, next boy, next girl)
and so on...

There is a sequence to obtaining kids and that needs to be factored in. Many ways will lead to 2 girls and 2 boys hence the probability of obtaining 2 girls and 2 boys will be higher than the probability of obtaining 4 girls. The question clearly tells us that we don't know which 2 are girls so we cannot say that this the case where first child and second child are girls and others are boys. Hence you need to follow the method discussed above: https://gmatclub.com/forum/ms-barton-has ... 58#p991399

It is similar to the coin flipping case. Every coin flip is independent of the other but the probability of obtaining 4 heads in 4 flips is less than the probability of obtaining 2 heads and 2 tails.

Why are we getting into arrangements. The question is about probability (no matter in which order). Then the answer should be 1/4. why is it wrong?

Ms. Barton has four children. You are told correctly that she has at least two girls but you are not told which two of her four children are those girls. What is the probability that she also has two boys? (Assume that the probability of having a boy is the same as the probability of having a girl.)

(A) 1/4
(B) 3/8
(C) 5/11
(D) 1/2
(E) 6/11

GMATGuruNY avigutman KarishmaB what would prompt us in this question to take permutations? We could very well have only taken gggb,gggg,ggbb as cases and considered probability to be 1/3. lets ignore that solutions dont have this as ans because its wrong but pls tell the reasoning here for doing so? In other words why would it be so that each of the three possibilities lead to different number of orders for each