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A set of pictures of butterflies contains 10 pictures. Jim bought 3 of

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A set of pictures of butterflies contains 10 pictures. Jim bought 3 of  [#permalink]

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New post 08 Sep 2016, 04:18
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A set of pictures of butterflies contains 10 pictures. Jim bought 3 of the pictures. 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?

A. 14/15
B. 3/5
C. 6/13
D. 7/15
E. 7/10

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Re: A set of pictures of butterflies contains 10 pictures. Jim bought 3 of  [#permalink]

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New post 08 Sep 2016, 06:19
1
Jim selects 3 pics.. So the 2 pics are to be selected from the remaining 7

2 pics to be selected from 10 (sample space) = 10C2 = 45

2 pics to be selected from 7 pics (i.e. pics not chosen by Mr. Jim) = 7C2 = 21

The required probability = 21/45 = 7/15

Option D
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Re: A set of pictures of butterflies contains 10 pictures. Jim bought 3 of  [#permalink]

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New post 09 Sep 2016, 01:42
total = 10c2 =45;
selecting those not selected by jim = 7c2 = 21
probability = 21/45 = 7/15
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Re: A set of pictures of butterflies contains 10 pictures. Jim bought 3 of  [#permalink]

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New post 11 Sep 2016, 00:01
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Bunuel wrote:
A set of pictures of butterflies contains 10 pictures. Jim bought 3 of the pictures. 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?

A. 14/15
B. 3/5
C. 6/13
D. 7/15
E. 7/10


Probability of first pick not one of the Jim's picture is = \(\frac{7}{10}\) (except for 3 which Jim purchased)
Probability of second pick not one of the Jim's picture is = \(\frac{6}{9}\)

Total = \(\frac{7}{10} * \frac{6}{9} = \frac{7}{15}\)

Answer is D
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Re: A set of pictures of butterflies contains 10 pictures. Jim bought 3 of  [#permalink]

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New post 11 Sep 2016, 05:57
Bunuel wrote:
A set of pictures of butterflies contains 10 pictures. Jim bought 3 of the pictures. 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?

A. 14/15
B. 3/5
C. 6/13
D. 7/15
E. 7/10


Since Jim has bought 3 of the pictures, the subset from which the 2 pictures should be chosen are the 7 pictures not bought by Jim from the universe of 10.

The first picture can be one of the 7 from the 10 with probability 7/10
The second picture can be one of the 6 from the 9 remaining with probability 6/9
The total probability will be 7/10 x 6/9. On cancellation, this comes to 7/15.

Thus, the answer is D - 7/15.
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Re: A set of pictures of butterflies contains 10 pictures. Jim bought 3 of  [#permalink]

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New post 02 Jul 2018, 10:09
Bunuel wrote:
A set of pictures of butterflies contains 10 pictures. Jim bought 3 of the pictures. 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?

A. 14/15
B. 3/5
C. 6/13
D. 7/15
E. 7/10


The number of ways to select 2 pictures from 10 is 10C2 = 10!/(2! x 8!) = (10 x 9)/2 = 45.

The number of ways to select 2 pictures that are not bought by Jim is 7C2 = (7 x 6)/2 = 21 (notice that 7 pictures are not bought by Jim if he bought 3 of the 10 pictures).

Thus, the probability is 21/45 = 7/15.

Alternate Solution:

The probability that the first chosen picture is not one of the ones bought by Jim is 7/10.

The probability that the second chosen picture is not one of the ones bought by Jim is 6/9.

The probability that neither of the chosen pictures is bought by Jim is 7/10 x 6/9 = 42/90 = 7/15.

Answer: D
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Re: A set of pictures of butterflies contains 10 pictures. Jim bought 3 of  [#permalink]

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New post 21 Jul 2018, 06:13
1
Bunuel wrote:
A set of pictures of butterflies contains 10 pictures. Jim bought 3 of the pictures. 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?

A. 14/15
B. 3/5
C. 6/13
D. 7/15
E. 7/10



Hello Team..........I might be finally wrong , but I am a bit confused at this moment :
Total # pictures =10 , # pictures Jim bought = 3 , Thus remaining pictures = 7

# ways selecting 2 out of 10 = 10C2= 10*9/2 = 45 = total # out comes
# ways selecting 2 out of 3 = 3C2= 3*2/2 = 3 = # out comes when both the pictures are selected from Jims's 3
# ways selecting 2 out of 7 = 7C2= 7*6/2 = 21= # out comes when none of the pictures are selected from Jims's 3...........all good untill now.

Now the Q-stem asks : what is the probability that both pictures are not those that were already bought by Jim?

To me the interpretation of both pictures are not those that were already bought by Jim is ............. In this selection
case 1: There could be 1 out of 2 that is part of Jim's selection & the other one is not a part of Jim's Selection (i.e., part of other 7)
case 2: None of these are part of Jim's selection (i.e., part of other 7)

In other words : 1 - P (when both the selection are part of Jim's selections)
= \(1- (\frac{{# ways selecting 2 out of 3}}{{# ways selecting 2 out of 10}})\)
= \(1- (\frac{{3}}{{45}})\)
= \((\frac{{42}}{{45}})\)
= \((\frac{{14}}{{15}})\)..................................Thus I am tending to vote for A....... But then I saw the other solutions. :( :(
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Re: A set of pictures of butterflies contains 10 pictures. Jim bought 3 of  [#permalink]

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New post 22 Jul 2018, 00:52
u1983 wrote:
Bunuel wrote:
A set of pictures of butterflies contains 10 pictures. Jim bought 3 of the pictures. 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?

A. 14/15
B. 3/5
C. 6/13
D. 7/15
E. 7/10



Hello Team..........I might be finally wrong , but I am a bit confused at this moment :
Total # pictures =10 , # pictures Jim bought = 3 , Thus remaining pictures = 7

# ways selecting 2 out of 10 = 10C2= 10*9/2 = 45 = total # out comes
# ways selecting 2 out of 3 = 3C2= 3*2/2 = 3 = # out comes when both the pictures are selected from Jims's 3
# ways selecting 2 out of 7 = 7C2= 7*6/2 = 21= # out comes when none of the pictures are selected from Jims's 3...........all good untill now.

Now the Q-stem asks : what is the probability that both pictures are not those that were already bought by Jim?

To me the interpretation of both pictures are not those that were already bought by Jim is ............. In this selection
case 1: There could be 1 out of 2 that is part of Jim's selection & the other one is not a part of Jim's Selection (i.e., part of other 7)
case 2: None of these are part of Jim's selection (i.e., part of other 7)

In other words : 1 - P (when both the selection are part of Jim's selections)
= \(1- (\frac{{# ways selecting 2 out of 3}}{{# ways selecting 2 out of 10}})\)
= \(1- (\frac{{3}}{{45}})\)
= \((\frac{{42}}{{45}})\)
= \((\frac{{14}}{{15}})\)..................................Thus I am tending to vote for A....... But then I saw the other solutions. :( :(



I also used the same approach initially.
Can anybody tell me where we are wrong?????????????
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A set of pictures of butterflies contains 10 pictures. Jim bought 3 of  [#permalink]

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New post Updated on: 22 Jul 2018, 17:53
Hello Experts……..Bunuel, chetan2u, EgmatQuantExpert, stonecold, abhimahna ……..could you please put some light on our confusion in this problem?
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Originally posted by u1983 on 22 Jul 2018, 08:58.
Last edited by u1983 on 22 Jul 2018, 17:53, edited 2 times in total.
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A set of pictures of butterflies contains 10 pictures. Jim bought 3 of  [#permalink]

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New post 22 Jul 2018, 17:33
I always struggle with combinatorics problems. Graphical approach can ease my "pain". I hope it will help someone too. See attached file.
Attachments

1.png
1.png [ 8.88 KiB | Viewed 394 times ]

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Re: A set of pictures of butterflies contains 10 pictures. Jim bought 3 of  [#permalink]

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New post 22 Jul 2018, 17:52
Hero8888 wrote:
I always struggle with combinatorics problems. Graphical approach can ease my "pain". I hope it will help someone too. See attached file.



Hi Hero8888

My point is that .... the Q-stem asks : both pictures are not those that were already bought by Jim
So out of 2 to be selected 1 could be selected from Jim's 3.

If It had asked none of the pics to be selected from jims 3.... then I completely agree with your reasoning.
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A set of pictures of butterflies contains 10 pictures. Jim bought 3 of  [#permalink]

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New post 22 Jul 2018, 18:12
u1983 wrote:
Hero8888 wrote:
I always struggle with combinatorics problems. Graphical approach can ease my "pain". I hope it will help someone too. See attached file.



Hi Hero8888

My point is that .... the Q-stem asks : both pictures are not those that were already bought by Jim
So out of 2 to be selected 1 could be selected from Jim's 3.

If It had asked none of the pics to be selected from jims 3.... then I completely agree with your reasoning.


I think that the question imposes straight restriction on those 3 that have been bought by Jim. So you were asked how can you choose 2 out of 10, so not to choose 3 similar to Jim's.
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Re: A set of pictures of butterflies contains 10 pictures. Jim bought 3 of  [#permalink]

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New post 22 Jul 2018, 18:44
7/10 times 6/9
=42/90
=7/15
How did 45% get the timer wrong??
Re: A set of pictures of butterflies contains 10 pictures. Jim bought 3 of &nbs [#permalink] 22 Jul 2018, 18:44
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