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We can treat this exactly as we would a coin flip question; think of it as:hongwang9703 wrote: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 wrote: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
I am trying to understand the logic but can't get it through the prob. tree.Stuart Kovinsky wrote:We can treat this exactly as we would a coin flip question; think of it as:hongwang9703 wrote: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!
"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 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 wrote:Abdulla,
I agree with your approach. I was wondering the same thing !
The problem with your solution is that there aren't 4! total possible outcomes.Abdulla wrote:I am trying to understand the logic but can't get it through the prob. tree.Stuart Kovinsky wrote:We can treat this exactly as we would a coin flip question; think of it as:hongwang9703 wrote: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!
"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
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??
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.Stuart Kovinsky wrote:The problem with your solution is that there aren't 4! total possible outcomes.Abdulla wrote:I am trying to understand the logic but can't get it through the prob. tree.Stuart Kovinsky wrote:We can treat this exactly as we would a coin flip question; think of it as:hongwang9703 wrote: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!
"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
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 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!.
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