A couple decides to have 4 children. lf 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
OA : A
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vinay1983
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Last edited by vinay1983 on Tue Aug 27, 2013 1:28 am, edited 1 time in total.
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Hi vinay1983,
The answer that you listed as the *correct answer* is incorrect.
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 = [spoiler]6/16 = 3/8 = A[/spoiler]
GMAT assassins aren't born, they're made,
Rich
The answer that you listed as the *correct answer* is incorrect.
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 = [spoiler]6/16 = 3/8 = A[/spoiler]
GMAT assassins aren't born, they're made,
Rich
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Deepthi Subbu
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Since every child is equally likely to be born, the probability of getting 2 girls and 2 boys is (1/2)^4 = 1/16.
Now the 2 girls and the two boys can be arranged in 4! ways. However G1G2B1B2 is the same as G2G1B1B2 and many other such combinations. So 4!/4(1/16) = 3/8.
Are you sure that OA is E?
Now the 2 girls and the two boys can be arranged in 4! ways. However G1G2B1B2 is the same as G2G1B1B2 and many other such combinations. So 4!/4(1/16) = 3/8.
Are you sure that OA is E?
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P(exactly n times) = P(one way) * total possible ways.vinay1983 wrote:A couple decides to have 4 children. lf 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
OA : A
P(one way):
One way to get an equal number of boys and girls is for first two children born to be boys and for the last two children born to be girls.
P(BBGG) = 1/2 * 1/2 * 1/2 * 1/2 = 1/16.
Total possible ways:
BBGG is only ONE WAY to get exactly 2 boys and exactly 2 girls.
Now we must account for ALL OF THE WAYS to get exactly 2 boys and exactly 2 girls.
Any arrangement of the letters BBGG will yield exactly 2 boys and 2 girls.
Thus, to account for ALL OF THE WAYS to get exactly 2 boys and 2 girls, the result above must be multiplied by the number of ways to arrange the letters BBGG.
Number of ways to arrange 4 elements = 4!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 2! to account for the two identical B's and by another 2! to account for the two identical G's:
4!/(2!2!) = 6.
Multiplying the results above, we get:
P(exactly 2 boys and 2 girls) = 6 * 1/16 = 3/8.
The correct answer is A.
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As Rich mentioned, we can use a table (or tree diagram) to solve this question. It may take a while, but not more than 2 minutes to do this. So, if you don't spot a faster approach, you should consider using a tree diagram.vinay1983 wrote:A couple decides to have 4 children. lf 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
As you can see, all 16 possible scenarios are listed here.
Of these 16 cases, 6 of them meet the "2 boys and 2 girls" criterion.
So, the probability = [spoiler]6/16 = 3/8 = A[/spoiler]
Cheers,
Brent
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ganeshrkamath
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Here, since I don't have to care about the order in which they are born,vinay1983 wrote:A couple decides to have 4 children. lf 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
OA : A
the favorable cases are 4C2 = 4*3/2 = 6
The sample space = 2^4 = 16
Required probability = 6/16 = 3/8
Choose A
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