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A couple decides to have 6 children

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A couple decides to have 6 children

by nahid078 » Thu Sep 01, 2016 6:58 am
A couple decides to have 6 children. If they succeed in having 6 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 4 girls and 2 boys?
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by GMATGuruNY » Thu Sep 01, 2016 7:06 am
nahid078 wrote:A couple decides to have 6 children. If they succeed in having 6 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 4 girls and 2 boys?
P(exactly n times) = P(one way) * total possible ways.

P(one way):
One way to get exactly 4 girls and 2 boys is for first four children born to be girls and for the last two children born to be boys.
P(GGGGBB) = 1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/64.

Total possible ways:
GGGGBB is only ONE WAY to get exactly 4 girls and 2 boys.
Now we must account for ALL OF THE WAYS to get exactly 4 girls and 2 boys.
Any arrangement of the letters GGGGBB will yield exactly 4 girls and 2 boys.
Thus, to account for ALL OF THE WAYS to get exactly 4 girls and 2 boys, the result above must be multiplied by the number of ways to arrange the letters GGGGBB.
Number of ways to arrange 6 elements = 6!.
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 4! to account for the four identical G's and by 2! to account for the two identical B's:
6!/(4!2!) = 15.

Multiplying the results above, we get:
P(exactly 4 girls and 2 boys) = 15 * 1/64 = [spoiler]15/64[/spoiler].

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by [email protected] » Thu Sep 01, 2016 10:10 am
Hi nahid078,

For future reference, you should post the ENTIRE prompt (including the 5 answer choices). With certain questions, the answer choices provide a clue as to how you might go about answering the question. Without having those answers, we're forced to do math (which might not be the fastest, nor the easiest, way to get to the answer).

Mitch's approach is spot-on, so I won't rehash that here. Given the specific situation, you could use 'brute force' to 'map out' all of the options though.

If we treat this as a permutation, since each of the 6 children has an equal chance of being a boy or a girl, there are (2)(2)(2)(2)(2)(2) = 64 outcomes. We're asked for exactly 4 girls and 2 boys. Those options would be:

GGGGBB
GGGBGB
GGBGGB
GBGGGB
BGGGGB

GGGBBG
GGBGBG
GBGGBG
BGGGBG

GGBBGG
GBGBGG
BGGBGG

GBBGGG
BGBGGG
BBGGGG

Total options = 15 out of 64

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by Matt@VeritasPrep » Thu Sep 01, 2016 4:43 pm
Another approach here:

Assuming that P(boy) = 1/2, we know that P(0 boys) = (1/2)� = 1/64.

If we want P(exactly 1 boy), we have (1/2)� * 6, since any of the six can be the one boy.

If we want P(exactly 2 boys), we have (1/2)� * (6 choose 2), since any two of the six can be the two boys.

Since P(exactly 4) = P(exactly 2), we simply do (1/2)� * (6 choose 2), or (1/2)� * (6*5)/2, or 15/64. (If that doesn't make sense, you could also do the more intuitive way: (1/2)� * (6 choose 4), which gives the same result.)
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