probability

[romitvsingh]

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

[Anurag@Gurome]

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.

[amit2k9]

(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.

[jaguar123]

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.

[pemdas]

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.

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.