Friday, September 07, 2007

Answer to Boys and Girls problem

The answer is .4

This was a very simple problem. The key in probability is to keep your thinking as simple as the problem allows one to.

There are 16 permutations for the four children.
2*2*2*2 = 16.
GGGG, BBBB, BBGG, BGBG, BGGB, GGBB, GBGB, GBBG, GBBB, BGBB, BBGB, BBBG, GGGB, GGBG, GBGG, BGGG
Now, since it is already given that one of them is a girl, the permutation of BBBB is ruled out. That leaves 15 permutations.
In that count the ones that have just one girl. That would be six. BBGG, BGBG, BGGB, GGBB, GBGB, GBBG

So the probability is 6/15 = 0.4

Of course in CAT, you should arrive at 6 and 15 using permutation formula to make the calculation quicker

2 Comments:

Blogger Atul said...

one issue i have with this solution is that it takes into account permutation instead of combination which is the correct approach according to me.

my solution:-

we require that at least one child should be girl.
so it is similar to 3 children which can be boys as wellas girls.
so the combinations will be 2girl 1 boy, 1 girl 2 boy, 3 boys, 3 girls.

i.e. 4 possibilities.

now total sets can be 4 boys, 4 girls, 1 boy 3 girl, 2 boy 2 girl , 3 boy 1 girl.
i.e. 5 possibilities.

so answer is .8

maybe the problem takes every boy and firl as non identical entities but the formulation of question looks like the order doesnt matter just the number.

10:50 PM, September 09, 2007  
Blogger Gyan-ee said...

Atul,

But the problem is that the cases you have described are not "equally likely". If someone has 4 children, the chances of 1 boy three girls is more likely than 4 girls.

The reason for this is that this one boy child has a chance of being born in any four places (permutation in simple words :))

It is similar to the fact that 5 heads and 25 tails are more likely than 30 tails in a fair coin.

2:45 AM, September 11, 2007  

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