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by romitvsingh » Mon Nov 28, 2011 8:08 pm
A couple decides to have 4 children. If they suceed in having 4 children & each is equally likely to be a boy or a girl, what is the probaility that the will have exactly 2 girls or 2 boys?

A) 3/8
B) 1/4
C) 3/16
D) 1/8
E) 1/16

Pls explain in detail
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by Anurag@Gurome » Mon Nov 28, 2011 8:38 pm
romitvsingh wrote:A couple decides to have 4 children. If they suceed in having 4 children & each is equally likely to be a boy or a girl, what is the probaility that the will have exactly 2 girls or 2 boys?

A) 3/8
B) 1/4
C) 3/16
D) 1/8
E) 1/16

Pls explain in detail
Possibilities of having 2 boys and 2 girls = BBGG, GGBB, BGBG, GBGB, BGGB, GBBG, which means 6 possibilities.

Total no. of ways in which 2 children can be born = 2^4 = 16
So, required probability = 6/16 = 3/8

The correct answer is A.
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by amit2k9 » Mon Nov 28, 2011 11:06 pm
(1/2)^4 * 4C2 = 3/8
4 kids with probability of 50 for a girl of a boy.
4C2 for exactly two boys or two girls.
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by jaguar123 » Sat Dec 03, 2011 12:22 am
One doubt.. what is the mistake with this method.

Total number of combinations - without oder -5.

1. 4B 0G - 0B,4G
2. 3B 1G - 1B 3G
3. 2B,2G

there is only one combination that works fine - 2B,2G - 1/5. Please help.
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by pemdas » Sat Dec 03, 2011 1:15 am
this q. can be simplified if we ask "When tossing a coin 4 times, how many times we have 2 tails and 2 heads exactly?" Try solving this way and it will be much easier.

Now with your doubt again, if you consider coin instead of child. Let me ask you how you can know if the coin tossed by you will land in 4 tails (or heads) and 0 heads (or tails), 3 tails (or heads) and 3 heads (or tails), ...?
You've prearranged the order of not "the only possible ways" but the order of "samples: 4B 0G, 4G 0B, 3B 1G, 3G 1B ... etc,"
Each sample set may have different probability and will be the baseline for a new question :)

stick with coins and try to think further about this question.
jaguar123 wrote:One doubt.. what is the mistake with this method.

Total number of combinations - without oder -5.

1. 4B 0G - 0B,4G
2. 3B 1G - 1B 3G
3. 2B,2G

there is only one combination that works fine - 2B,2G - 1/5. Please help.
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