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?

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)

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