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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?
(A) 3/8
(B) 1/4
(C) 3/16
(D) 1/8
(E) 1/16
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(A) 3/8
(B) 1/4
(C) 3/16
(D) 1/8
(E) 1/16
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SOLUTION
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?
(A) 3/8
(B) 1/4
(C) 3/16
(D) 1/8
(E) 1/16
\(P(GGBB)=\frac{4!}{2!2!}*(\frac{1}{2})^4=\frac{3}{8}\), we should mulitply by \(\frac{4!}{2!2!}\) as the scenario GGBB can occur in several ways: GGBB, BBGG, GBGB, ... # of scenarios possible is # of permutations of 4 letters GGBB out of which 2 G's and 2 B's are identical and equals to \(\frac{4!}{2!2!}\).
Answer: A.
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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?
(A) 3/8
(B) 1/4
(C) 3/16
(D) 1/8
(E) 1/16
\(P(GGBB)=\frac{4!}{2!2!}*(\frac{1}{2})^4=\frac{3}{8}\), we should mulitply by \(\frac{4!}{2!2!}\) as the scenario GGBB can occur in several ways: GGBB, BBGG, GBGB, ... # of scenarios possible is # of permutations of 4 letters GGBB out of which 2 G's and 2 B's are identical and equals to \(\frac{4!}{2!2!}\).
Answer: A.
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21. Combinatorics/Counting Methods
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Math: Combinatorics
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no of ways of getting P(GGBB) is 4!/2!*2!;
Total no of ways is 2^n =2^4 =16;
6/16 = 3/8;
We can consider this question to a coin that is flipped for 4 times . what is the probability of getting exactly two heads .
P(all out comes) = 1/2 *1/2 *1/2 *1/2 =1/16;
P(favorable outcomes) = 4!/(2! * 2!) = 6/16 =3/8;
(OR)
Second Approach
GBGB
GGBB
BBGG
BGBG
GBBG
BGGB
6 possible ways .
total no of ways is
Baby can be a boy or a girl.
For each baby the probability is 1/2 ;for 4 babies it's 1/16;
6/16 = 3/8;
Total no of ways is 2^n =2^4 =16;
6/16 = 3/8;
We can consider this question to a coin that is flipped for 4 times . what is the probability of getting exactly two heads .
P(all out comes) = 1/2 *1/2 *1/2 *1/2 =1/16;
P(favorable outcomes) = 4!/(2! * 2!) = 6/16 =3/8;
(OR)
Second Approach
GBGB
GGBB
BBGG
BGBG
GBBG
BGGB
6 possible ways .
total no of ways is
Baby can be a boy or a girl.
For each baby the probability is 1/2 ;for 4 babies it's 1/16;
6/16 = 3/8;
NoHalfMeasures
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By fundamental counting principle,
Total No of out comes: 2^4 = 16
Total Desired outcomes: No of ways to arrange BBGG by MISSISSIPPI rule= 4!/(2!*2!) = 6
Probability is 6/16 = 3/8
Total No of out comes: 2^4 = 16
Total Desired outcomes: No of ways to arrange BBGG by MISSISSIPPI rule= 4!/(2!*2!) = 6
Probability is 6/16 = 3/8
So here we have four positions and there are two options to fill each position
so total number of cases=2x2x2x2=16
now we need 2boys and 2 girls OR we can say that we simply need 2 boys because if its not a boy it has to be a girl
favourable cases=4C2=6
probability=6/16=3/8
(A)
so total number of cases=2x2x2x2=16
now we need 2boys and 2 girls OR we can say that we simply need 2 boys because if its not a boy it has to be a girl
favourable cases=4C2=6
probability=6/16=3/8
(A)
Posts: 577
Bunuel
Let us handle a single case. First boy, second boy, third girl, fourth girl. The probability of this case is 1/2*1/2*1/2*1/2 = 1/16
Two boys or two girls can be born among 4 children in 4C2 ways= 6 ways
So the required probability is: total number of ways * probability of one such way= 6* (1/16)=3/8
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Srinivasan Vaidyaraman
General Discussion
1/2 is the chance of boy or girl
1/2 * 1/2 * 1/2 * 1/2 = 1/16 (4 children, boy or girl)
Possible ways of 2 boys, 2 girls:
GGBB
BBGG
GBGB
BGBG
BGGB
GBBG
= 6 ways
we need "OR" --> 1/16 OR 1/16 ....
1/16 + 1/16 + 1/16 ....... +1/16 = 6/16 = 3/8
Answer A
1/2 * 1/2 * 1/2 * 1/2 = 1/16 (4 children, boy or girl)
Possible ways of 2 boys, 2 girls:
GGBB
BBGG
GBGB
BGBG
BGGB
GBBG
= 6 ways
we need "OR" --> 1/16 OR 1/16 ....
1/16 + 1/16 + 1/16 ....... +1/16 = 6/16 = 3/8
Answer A
Why are we ordering here? I mean, I would think BBGG to be the same as BGBG - since the question asks about exactly 2 girls and 2 boys, irrespective of any order of delivery.
We have 5 possibilities: 4 boys, 3 boys and 1 girl, 2 boys and 2 girls, 1 boy and 3 girls, and finally all 4 girls.
So shouldn't the probability be 1/5? too simplistic - I know. but where am I wrong?
We have 5 possibilities: 4 boys, 3 boys and 1 girl, 2 boys and 2 girls, 1 boy and 3 girls, and finally all 4 girls.
So shouldn't the probability be 1/5? too simplistic - I know. but where am I wrong?
Binomial Probability
The probability of achieving exactly k successes in n trials is shown below.
Formula: P(Probability of K successes in n trials) = nCk p^k q^n-k
n = number of trials
k = number of successes
n – k = number of failures
p = probability of success in one trial
q = 1 – p = probability of failure in one trial
According to our question
n(4 children) = 4
k( we want exactly 2 girls) = 2
n – k = 2
p (probability of getting a girl in one trial) = 1/2
q = 1 – p = 1/2
4C2 * 1/2^2 * 1/2^2 =3/8
Binomial can also be used for problem like coins.
Binomial is used when following conditions are satisfied.
Fixed number of trials
Independent trials
Two different classifications
Probability of success stays the same for all trials
The probability of achieving exactly k successes in n trials is shown below.
Formula: P(Probability of K successes in n trials) = nCk p^k q^n-k
n = number of trials
k = number of successes
n – k = number of failures
p = probability of success in one trial
q = 1 – p = probability of failure in one trial
According to our question
n(4 children) = 4
k( we want exactly 2 girls) = 2
n – k = 2
p (probability of getting a girl in one trial) = 1/2
q = 1 – p = 1/2
4C2 * 1/2^2 * 1/2^2 =3/8
Binomial can also be used for problem like coins.
Binomial is used when following conditions are satisfied.
Fixed number of trials
Independent trials
Two different classifications
Probability of success stays the same for all trials
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donkadsw
Bump seeking an answer to the question above:
I got 1/5 as well.
BBBB
BBBG
BBGG
GGGB
GGGG
1/5 outcomes
The question doesn't say anything about the order. How do we know when order matters?
Dondarrion
In this case we are saying that the probability of BGGG is the same as the probability of BBGG. But that is not so.
You can have 1 boy and 3 girls in 4 ways: BGGG, GBGG, GGBG, GGGB
But you can have 2 boys and 2 girls in 6 ways: BBGG, BGGB, GGBB, BGBG, GBGB, GBBG
So the probability depends on the number of ways in which you can get 2 boys and 2 girls.
Think of it this way: if you throw two dice, is the probability of getting a sum of 2 same as the probability of getting a sum of 8? No.
For sum fo 2, you must get 1 + 1 only.
For sum of 8, you could get 4 + 4 or 3 + 5 or 2 + 6 etc.
So probability of getting sum of 8 would be higher.
In the same way, the order matters in this question.
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Hi guys, I don't quite understand why we square 4 instead of factorial of 4. Could you please help?
ManonZ
Do you mean in the total number of cases? If yes, then it is actually 2^4.
First child can happen in 2 ways (boy or girl).
Second, third and fourth kids can also happen in 2 ways.
Total number of ways = 2*2*2*2 = 16
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EMPOWERgmatRichC
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Expert reply
Hi All,
There are a couple of ways to answer this question:
1) You could make a table of all the options (there are only 16 possible outcomes) and count up all the ways to get 2 boys and 2 girls
or
2) You can do the math
Here's the math approach:
Since each child has an equal chance of ending up as a boy or girl, there are 2^4 possibilities = 16 possibilities
It also doesn't matter which 2 of the 4 children are boys, so you can treat this part of the question as a combination formula question...
4c2 = 4!/[2!2!] = 6 ways to get 2 boys and 2 girls
Final Answer =
GMAT assassins aren't born, they're made,
Rich
There are a couple of ways to answer this question:
1) You could make a table of all the options (there are only 16 possible outcomes) and count up all the ways to get 2 boys and 2 girls
or
2) You can do the math
Here's the math approach:
Since each child has an equal chance of ending up as a boy or girl, there are 2^4 possibilities = 16 possibilities
It also doesn't matter which 2 of the 4 children are boys, so you can treat this part of the question as a combination formula question...
4c2 = 4!/[2!2!] = 6 ways to get 2 boys and 2 girls
Final Answer =
6/16 = 3/8 = A
GMAT assassins aren't born, they're made,
Rich
JeffTargetTestPrep
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Expert reply
Posts: 2,973
Bunuel
We need to determine the probability of P(B-B-G-G).
P(B-B-G-G) = (1/2)^4 = 1/16
The number of ways to arrange B-B-G-G is 4C2 = 4!/(2! x 2!) = 3 x 2 = 6.
Thus, the total probability is 6/16 = 3/8.
Answer: A
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I thought that order does not matter for when child is born, therefore
Total number of events = 5 (explained below)
0 Boys 4 Girls
1 Boy 3 Girls
2 Boys 2 Girls
3 Boys 1 Girl
4 Boys 0 Girl
Probability (2 B and 2 G exactly) = 1/5
Not sure where I am assuming and making a mistake.
Total number of events = 5 (explained below)
0 Boys 4 Girls
1 Boy 3 Girls
2 Boys 2 Girls
3 Boys 1 Girl
4 Boys 0 Girl
Probability (2 B and 2 G exactly) = 1/5
Not sure where I am assuming and making a mistake.
EMPOWERgmatRichC
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Expert reply
shavarna
Hi shavarna,
The 5 options you've listed are NOT all equally likely, so you would have to do a bit more work to get to the correct answer. There are a couple of ways to answer this question:
1) You could make a table of all the options (since each child could be a boy or a girl, there are only 2^4 = 16 possible outcomes) and determine all the ways to get 2 boys and 2 girls
or
2) You can do the math
Here's the math approach:
Since each child has an equal chance of ending up as a boy or girl, there are 2^4 possibilities = 16 possibilities. It also doesn't matter which 2 of the 4 children are boys, so you can treat this part of the question as a combination formula question...
4c2 = 4!/[2!2!] = 6 ways to get 2 boys and 2 girls out of 16 possibilities. 6/16 = 3/8
Final Answer =
GMAT assassins aren't born, they're made,
Rich
Posts: 137
EMPOWERgmatRichC
Hi EMPOWERgmatRichC, chetan2u,
But that's also valid method right as nowhere stated in question that in how many different way this is possible.
EMPOWERgmatRichC
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Expert reply
Hi Gmatprep550,
The question asks for the PROBABILITY that exactly 2 of the 4 children are boys, so the total number of possible outcomes IS a factor in this question.
If you describe the 5 possible outcomes as...
0 Boys 4 Girls
1 Boy 3 Girls
2 Boys 2 Girls
3 Boys 1 Girl
4 Boys 0 Girl
...then you might think that there is a 1/5 chance of having 2 boys and 2 girls, but that is NOT mathematically correct (since the 5 possible outcomes do NOT all have an equal likelihood of occurring).
GMAT assassins aren't born, they're made,
Rich
The question asks for the PROBABILITY that exactly 2 of the 4 children are boys, so the total number of possible outcomes IS a factor in this question.
If you describe the 5 possible outcomes as...
0 Boys 4 Girls
1 Boy 3 Girls
2 Boys 2 Girls
3 Boys 1 Girl
4 Boys 0 Girl
...then you might think that there is a 1/5 chance of having 2 boys and 2 girls, but that is NOT mathematically correct (since the 5 possible outcomes do NOT all have an equal likelihood of occurring).
GMAT assassins aren't born, they're made,
Rich
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