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Probability - class of children

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Probability - class of children

by Brent@GMATPrepNow » Sat Jan 31, 2009 2:48 pm
A certain class has an equal number of boys and girls. If three children are chosen at random, the probability that all three children are the same GMAT is 1/5. How many children are in the class?

A) 16
B) 18
C) 22
D) 24
E) 26

Please note that this is not an official GMAT question; it’s my attempt to create difficult (650+ level) GMAT-style questions for this forum.
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Re: Probability - class of children

by logitech » Sat Jan 31, 2009 3:08 pm
Boys= X
Girls = X

the probability that all three children are girls = 1/2 x 1/5 = 1/10

C(X:3)/C(2x:3) = 1/10

[x(x-1)(x-2)]/[(2x)(2x-1)(2x-2)] = 1/10

(x-2)/(8x-4) = 1/10

10X-20=8x-4

2x = 16
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Re: Probability - class of children

by Brent@GMATPrepNow » Sat Jan 31, 2009 3:14 pm
logitech wrote:Boys= X
Girls = X

the probability that all three children are girls = 1/2 x 1/5 = 1/10

C(X:3)/C(2x:3) = 1/10

[x(x-1)(x-2)]/[(2x)(2x-1)(2x-2)] = 1/10

(x-2)/(8x-4) = 1/10

10X-20=8x-4

2x = 16
Nice work, Logitech

Here's my soln:

Let k be the number of boys. This means that there are k girls and there are 2k children altogether.
Rather than work with both possible cases (all three are boys and all three are girls), it might be easiest to work with one case, say all 3 are girls.

If P(all 3 are same GMAT)=1/5, then P(all three are boys) + P(all three are girls) = 1/5
Since P(all three are boys) = P(all three are girls), then P(all three are girls) = 1/10
P(3 girls are selected) = P(1st child is a girl AND 2nd child is a girl AND 3rd child is a girl)
P(3 girls are selected) = P(1st child is a girl) x P(2nd child is a girl) x P(3rd child is a girl)
P(3 girls are selected) = k/2k x (k-1)/(2k-1) x (k-2)/(2k-2)
P(3 girls are selected) = (k-2)/(8k-4) (simplify)
We are told that P(3 girls are selected) = 1/10
So, we get the equation 1/10 = (k-2)/(8k-4)
Solve for k to get k=8
So, there are 16 children (answer choice A)
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by gaggleofgirls » Sat Jan 31, 2009 7:33 pm
I approached it differently:

X = Total number of children.

The probability that the first child chosen is of one GMAT = x/x = 1

The probability that the second child chosen is the same GMAT as the first = (x/2 -1)/x Because x/2 is all the chhildren of one GMAT and you already used one of them in the first choice.

The probability that the third child chosen matches the same GMAT as the first two = (x/2 -2)/x

Multiply the probabilities together (you can drop the first one, since it = 1) =
(x/2-1)(x/2-2) / (x-1)(x-2) = 1/5

At this point, seems easier just to plug in the answers to try than to work out the math...

(16/2 -1)(16/2-2) / (16-1)(16-2) = 7 * 6 / 15 * 14 = 7 * 3 * 2 / 15 * 7 * 2 = 3/15 = 1/5

Answer = A

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by raptor84 » Sat Jan 31, 2009 9:07 pm
backsolving....

let number of children be 16.

Boys = Girls = 8

Now number of ways 3 children can be selected is 3C16

= 16*15*14/3*2*1 = 8*5*14

Now the number of ways 3 boys can be selected from the 8 boys is

= 8*7*6/3*2*1 = 8*7

Therefore the probability that 3 children selected from the total would all be boys

= 8*7/8*5*14 = 1/10

Similarly solving for number of girls we get = 1/10

Total probability = 1/10+1/10 = 1/5 as provided in the question stem.
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