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Stat Question

by DCman401 » Wed Mar 21, 2012 6:50 am
A box of nine gloves contains two left-handed gloves and seven right-handed gloves. If two gloves are randomly selected from the box, without replacement (the first glove is not returned to the box after it is selected), what is the probability that there will be one right-handed glove and one left handed glove selected?

I'm in disagreement with a friend about what the correct answer is. Let me know what you think. Thanks
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by kmittal82 » Wed Mar 21, 2012 7:00 am
P(right handed glove) = 7/9
P(left handed glove) = 2/8

So, probability of 1 right handed glove and 1 left handed glove = (1/4) x (7/9) = 7/36

Whats the OA (and disagreement)? :)
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by DCman401 » Wed Mar 21, 2012 7:20 am
kmittal82 wrote:P(right handed glove) = 7/9
P(left handed glove) = 2/8

So, probability of 1 right handed glove and 1 left handed glove = (1/4) x (7/9) = 7/36

Whats the OA (and disagreement)? :)
The disagreement was in the fastest way to apprach the problem.

1. Probability of one right then one left. (7/9)*(2/8)
2. Probability of one left then one right. (2/9)*(7/8)

Both answers are 14/72 or 2/36.

I said for this problem you could stop there. My friend thought you needed to say the chance of drawing a right handed glove first is 7/9 and the chance of drawing a left handed glove first is 2/9 so therefore you needed to take (14/72)*7/9 + (14/72)*2/9. They both give you the same answer, and for practice purposes his way is good to do. But for the gmat when time is a factor, the first way I thought was better. I just wanted to see the way in which the first person explained their answer. Thank you.
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by Brent@GMATPrepNow » Wed Mar 21, 2012 7:30 am
DCman401 wrote:A box of nine gloves contains two left-handed gloves and seven right-handed gloves. If two gloves are randomly selected from the box, without replacement (the first glove is not returned to the box after it is selected), what is the probability that there will be one right-handed glove and one left handed glove selected?
I like to begin by rewriting the probability.

P(one of each) = P(1st selection left and 2nd selection right OR 1st selection right and 2nd selection left)
= P(1st selection left and 2nd selection right) + P(1st selection right and 2nd selection left)
= (2/9)(7/8) + (7/9)(2/8)
= 14/72 + 14/72
= 28/72
= 7/18

Cheers,
Brent
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by ronnie1985 » Wed Mar 21, 2012 7:47 am
Either first glove is right and second glove left or vice-versa. The probability is (2C1*7C1)/(9C1*8C1)+(7C1*2C1)/(9C1*8C1) = 2*7*2 / (8*9) = 7/18
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by kmittal82 » Wed Mar 21, 2012 8:41 am
Ah, thanks guys.. looks like I didn't consider all the possibilities
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by vikram4689 » Wed Apr 11, 2012 1:21 am
Brent how did you decide whether you need to ARRANGE and not CHOOSE because ultimately we are concerned for 1 left and 1 right so why does order matter..
Brent@GMATPrepNow wrote:
DCman401 wrote:A box of nine gloves contains two left-handed gloves and seven right-handed gloves. If two gloves are randomly selected from the box, without replacement (the first glove is not returned to the box after it is selected), what is the probability that there will be one right-handed glove and one left handed glove selected?
I like to begin by rewriting the probability.

P(one of each) = P(1st selection left and 2nd selection right OR 1st selection right and 2nd selection left)
= P(1st selection left and 2nd selection right) + P(1st selection right and 2nd selection left)
= (2/9)(7/8) + (7/9)(2/8)
= 14/72 + 14/72
= 28/72
= 7/18

Cheers,
Brent
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