Complex OG problem

[hongwang9703]

A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?

please solve if you are able and an explanation would be much appreciated!

[adam15]

this is a binomial problemjust use B(1/2,4)
with (2c2(1/2)^2(1/2)^2) +2c2(1/2)^2(1/2)^2=1/8

I hope this could be the right answer

[Stuart@KaplanGMAT]

A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?

please solve if you are able and an explanation would be much appreciated!

We can treat this exactly as we would a coin flip question; think of it as:

"If a fair coin is flipped 4 times, what's the probability of getting exactly 2 heads?"

There's an easy to use formula for coin flip (and pseudo-coin flip) questions:

Probability of getting exactly k results out of n flips = nCk/2^n

(nCk = n!/k!(n-k)!, the combinations formula.)

In this question, we have 4 flips and we want 2 heads, so n=4 and k=2:

4C2/2^4 = (4!/2!2!)/16 = (24/4)/16 = 6/16 = 3/8

[hongwang9703]

this is a binomial problemjust use B(1/2,4)
with (2c2(1/2)^2(1/2)^2) +2c2(1/2)^2(1/2)^2=1/8

I hope this could be the right answer

sorry 1/8 is inccorect, but genius mr.stuart solved it perfectly.

ALL HAILL Stuart!

how do u do it sir.

[Abdulla]

A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?

please solve if you are able and an explanation would be much appreciated!

We can treat this exactly as we would a coin flip question; think of it as:

"If a fair coin is flipped 4 times, what's the probability of getting exactly 2 heads?"

There's an easy to use formula for coin flip (and pseudo-coin flip) questions:

Probability of getting exactly k results out of n flips = nCk/2^n

(nCk = n!/k!(n-k)!, the combinations formula.)

In this question, we have 4 flips and we want 2 heads, so n=4 and k=2:

4C2/2^4 = (4!/2!2!)/16 = (24/4)/16 = 6/16 = 3/8

I am trying to understand the logic but can't get it through the prob. tree.
BBGG
GGBB
GBGB
BGBG
BGGB
GBBG
so these are the only possible ways which is 4!/2!2! = 6

the total possible outcomes 4! = 24, so 6/24 is 1/4 ?? what's wrong with my approach??

[heshamelaziry]

Abdulla,

I agree with your approach. I was wondering the same thing !

[Abdulla]

Abdulla,

I agree with your approach. I was wondering the same thing !

I hate Prob, Comb, and Permutations. The more I study, the more I got confused. I really don't know how to fix this issue.

[heshamelaziry]

Abdulla,

I am deeply suffering with all those probabilities, combination, .....etc. It is a fiasco. I am trying to learn them by trial and error and experience; meaning that when I see aproblem, i try to remember if I solved similar one and how i solved it. This is by all means a bad way to learn those MF, but i see no other road.

[papgust]

Guys,

don't worry too much about Probability, P&C. You hardly a question or two from these topics. So, if you find it very difficult to learn, don't dwell too much on the difficult problems. Just familiarize yourselves with easy problems and get on with other topics.

[Stuart@KaplanGMAT]

A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?

please solve if you are able and an explanation would be much appreciated!

We can treat this exactly as we would a coin flip question; think of it as:

"If a fair coin is flipped 4 times, what's the probability of getting exactly 2 heads?"

There's an easy to use formula for coin flip (and pseudo-coin flip) questions:

Probability of getting exactly k results out of n flips = nCk/2^n

(nCk = n!/k!(n-k)!, the combinations formula.)

In this question, we have 4 flips and we want 2 heads, so n=4 and k=2:

4C2/2^4 = (4!/2!2!)/16 = (24/4)/16 = 6/16 = 3/8

I am trying to understand the logic but can't get it through the prob. tree.
BBGG
GGBB
GBGB
BGBG
BGGB
GBBG
so these are the only possible ways which is 4!/2!2! = 6

the total possible outcomes 4! = 24, so 6/24 is 1/4 ?? what's wrong with my approach??

The problem with your solution is that there aren't 4! total possible outcomes.

The first child can be either a girl or a boy, as can the second, third and fourth children.

Since each child can be either a girl or a boy, there are 2*2*2*2 = 16 possible outcomes.

In a binomial distribution, there are 2^n possible outcomes, not n!.

[adam15]

this is a binomial problemjust use B(1/2,4)
with (2c4(1/2)^2(1/2)^2) +2c4(1/2)^2(1/2)^2=3/8

I made a small mistake, instead of 2,I should put 4 and you can check it yourself through the binomial
I hope this could be the right answer

[hongwang9703]

yeah I totaly agree with you guys on the Comb, Permutation and this coin-flip questions. I have spent hours and hours on this subject and I am dumbfounded when I come across a few medium difficulty questions, which is just very confusing for I have no idea which formula to use nor what number to plug in. But horray for the fact that there are only going to be a few of these questions on the gmat.

[hongwang9703]

wow after a little familiarity with this question, its actually very straight forward and simple....why didnt i get it before? I have come across questions like these many many times before. oh man, long way to go for me to get anywhere near 700...

[Abdulla]

A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?

please solve if you are able and an explanation would be much appreciated!

We can treat this exactly as we would a coin flip question; think of it as:

"If a fair coin is flipped 4 times, what's the probability of getting exactly 2 heads?"

There's an easy to use formula for coin flip (and pseudo-coin flip) questions:

Probability of getting exactly k results out of n flips = nCk/2^n

(nCk = n!/k!(n-k)!, the combinations formula.)

In this question, we have 4 flips and we want 2 heads, so n=4 and k=2:

4C2/2^4 = (4!/2!2!)/16 = (24/4)/16 = 6/16 = 3/8

I am trying to understand the logic but can't get it through the prob. tree.
BBGG
GGBB
GBGB
BGBG
BGGB
GBBG
so these are the only possible ways which is 4!/2!2! = 6

the total possible outcomes 4! = 24, so 6/24 is 1/4 ?? what's wrong with my approach??

The problem with your solution is that there aren't 4! total possible outcomes.

The first child can be either a girl or a boy, as can the second, third and fourth children.

Since each child can be either a girl or a boy, there are 2*2*2*2 = 16 possible outcomes.

In a binomial distribution, there are 2^n possible outcomes, not n!.

Thanks Stuart, At first I got the same total outcomes which is 16 but then I thought since I used the combinations formula to find out the desire outcomes then I must use the same formula to find the total outcomes, but now it's clear.