A group of pictures of butterflies contains 10 pictures. Jim had bought 3 of them. If 2 pictures are to be picked out from the 10 pictures, what is the probability that both pictures are not those that were already bought by Jim?
14/15
3/5
6/13
7/15
7/10
[spoiler]OA D
I did: Total number of ways to pick 2: 10C2 = 45
Total number of ways to pick the 2 of Jim's 3: 3C2 = 3
Therefore probability of picking the others = 1 - 3/45 = 42/45 = 14/15. Where did I go wrong?[/spoiler]
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A group of pictures of butterflies contains 10 pictures
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vishalpathak
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ganeshrkamath
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Favorable cases:vishalpathak wrote:A group of pictures of butterflies contains 10 pictures. Jim had bought 3 of them. If 2 pictures are to be picked out from the 10 pictures, what is the probability that both pictures are not those that were already bought by Jim?
14/15
3/5
6/13
7/15
7/10
[spoiler]OA D
I did: Total number of ways to pick 2: 10C2 = 45
Total number of ways to pick the 2 of Jim's 3: 3C2 = 3
Therefore probability of picking the others = 1 - 3/45 = 42/45 = 14/15. Where did I go wrong?[/spoiler]
Select 2 out of 7 not selected by Jim = 7C2
Sample space:
Select 2 out of any of the 10 pictures = 10C2
Probability = 7C2/10C2
= (7*6)/(10*9)
= 7/15
Choose D
Cheers
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Kelley School of Business (Class of 2016)
GMAT Score: 750 V40 Q51 AWA 5 IR 8
https://www.beatthegmat.com/first-attemp ... tml#688494
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ganeshrkamath
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I think you are not including some cases:vishalpathak wrote:A group of pictures of butterflies contains 10 pictures. Jim had bought 3 of them. If 2 pictures are to be picked out from the 10 pictures, what is the probability that both pictures are not those that were already bought by Jim?
14/15
3/5
6/13
7/15
7/10
[spoiler]OA D
I did: Total number of ways to pick 2: 10C2 = 45
Total number of ways to pick the 2 of Jim's 3: 3C2 = 3
Therefore probability of picking the others = 1 - 3/45 = 42/45 = 14/15. Where did I go wrong?[/spoiler]
Total number of ways of picking 1 out of Jim's and one not: 3C1 * 7C1 = 3*7 = 21
Therefore probability of picking the others = 1 - 3/45 - 21/45 = 21/45 = 7/15
Hope this helps.
Cheers
Every job is a self-portrait of the person who did it. Autograph your work with excellence.
Kelley School of Business (Class of 2016)
GMAT Score: 750 V40 Q51 AWA 5 IR 8
https://www.beatthegmat.com/first-attemp ... tml#688494
Kelley School of Business (Class of 2016)
GMAT Score: 750 V40 Q51 AWA 5 IR 8
https://www.beatthegmat.com/first-attemp ... tml#688494
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vishalpathak
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got it .. thanksganeshrkamath wrote:Favorable cases:vishalpathak wrote:A group of pictures of butterflies contains 10 pictures. Jim had bought 3 of them. If 2 pictures are to be picked out from the 10 pictures, what is the probability that both pictures are not those that were already bought by Jim?
14/15
3/5
6/13
7/15
7/10
[spoiler]OA D
I did: Total number of ways to pick 2: 10C2 = 45
Total number of ways to pick the 2 of Jim's 3: 3C2 = 3
Therefore probability of picking the others = 1 - 3/45 = 42/45 = 14/15. Where did I go wrong?[/spoiler]
Select 2 out of 7 not selected by Jim = 7C2
Sample space:
Select 2 out of any of the 10 pictures = 10C2
Probability = 7C2/10C2
= (7*6)/(10*9)
= 7/15
Choose D
Cheers
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Brent@GMATPrepNow
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The wording of this question is somewhat ambiguous.vishalpathak wrote:A group of pictures of butterflies contains 10 pictures. Jim had bought 3 of them. If 2 pictures are to be picked out from the 10 pictures, what is the probability that both pictures are not those that were already bought by Jim?
14/15
3/5
6/13
7/15
7/10
If we say that "both pictures are not those that were already bought by Jim," does this mean that NEITHER picture can be pre-purchased, or does it mean that it's okay for 1 picture to be pre-purchased (but not both)?
For example, if I forbid a child to eat BOTH of the candies in the bowl, does this mean that the child can eat one of the candies? Or does it mean that the child can eat NEITHER?
vishalpathak's answer of [spoiler]14/15[/spoiler] is correct if we say that ONE (but not both) of the pictures can be pre-purchased.
ganeshrkamath's answer of [spoiler]7/15[/spoiler] is correct if we say that NEITHER of the pictures can be pre-purchased.
Cheers,
Brent
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Hi vishalpathak,
ganeshrkamath has one way to do the "math" behind this question. Here's another.
Total Pictures = 10
Jim's = 3
Not Jim's = 7
Probability of picking 2 pictures that are NOT Jim's:
Probability of 1st: 7/10
Probability of 2nd: 6/9
(7/10)(6/9) = 42/90 = 7/15
Final Answer: D
GMAT assassins aren't born, they're made,
Rich
ganeshrkamath has one way to do the "math" behind this question. Here's another.
Total Pictures = 10
Jim's = 3
Not Jim's = 7
Probability of picking 2 pictures that are NOT Jim's:
Probability of 1st: 7/10
Probability of 2nd: 6/9
(7/10)(6/9) = 42/90 = 7/15
Final Answer: D
GMAT assassins aren't born, they're made,
Rich
