Assume that there are 4 Ms. Barton's children in a row: 1-2-3-4. Let show all possible combinations of them being girl(G) or boy(B) giving that there at least two girls. We are seeing each child 1,2,3,4 a s a unique(so, we distinguish between them)
BBGG
BGBG
GBBG
BGGB
GBGB
GGBB
BGGG
GBGG
GGBG
GGGB
GGGG

Overall 11 variants. 6 of them satisfy the condition of 2 boys exactly (they are highlighted in bold)
Therefore the probability is 6/11
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You need to determine the number of favourable arrangements (i.e. arrangements for which # girls = 2) out of all possible arrangements (i.e. arrangements for which # girls >= 2)

# arrangements for 2 girls and 2 boys (favourable arrangement) = 4! / (2!*2!) = 6
# arrangements for 3 girls and 1 boy = 4! / (3!*1!) = 4
# arrangements for 4 girls and 0 boys = 4! / 4! = 1

Thus, # favourable arrangements / # possible arrangements = 6 / (6 + 4+ 1) = 6/11

I like it too, and can be simplified even further.

There are normally 2*2*2*2=16 combinations.

We are told it is not 4B0G, and not 3B1G. It is clear that these are 1 combination, and 4 combinations (the G could be in either of 4 spots) respectively. So 16-5=11 combinations remain.

Since it is symmetric (p(g)=p(b)=0.5), there are also 5 combinations of 4G0B/3G1B, leaving 6/11.

p.s. Dont know if this is better, I never learnt all the equations with the '!' in them and when to use each, and so I always solve these intuitively.

OA IS E. Thanks guys for your help

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)
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Thanks VeritasPrepKarisma, for simplifying it so much.
Kudos for the details!