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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Please let me know if I am going in wrong direction.

Thanks in appreciation.