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A group of pictures of butterflies contains 10 pictures

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A group of pictures of butterflies contains 10 pictures

by vishalpathak » Wed Nov 20, 2013 6:12 am
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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by ganeshrkamath » Wed Nov 20, 2013 6:23 am
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]
Favorable cases:
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

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by ganeshrkamath » Wed Nov 20, 2013 6:26 am
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]
I think you are not including some cases:
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
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by vishalpathak » Wed Nov 20, 2013 6:42 am
ganeshrkamath wrote:
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]
Favorable cases:
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
got it .. thanks
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by Brent@GMATPrepNow » Wed Nov 20, 2013 6:59 am
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
The wording of this question is somewhat ambiguous.

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.

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Brent
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by [email protected] » Wed Nov 20, 2013 11:40 am
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

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