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

Question

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

Show Answer

Official Answer

D

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

ntngocanh19 · Answer

The numbers of way to pick 2 pic from 10 pic is C(10,2) = 45
The number of way to pick 2 pic from remain 7 pic not bought by Jim is C(7,2) = 21
=> The prob = 21/45 = 7/15
 Hence choice D

RamseyGooner · Answer

P(Not Jims Pictures) = Favourable Outcomes (Not Jims pictures)/ total Outcomes (All pictures) = 7C2/10C2 = 7/15

Hope this helps!

CrackverbalGMAT · Answer

Random experiment – picking 2 pictures at random from a set of 10 pictures.
Therefore, sample space = 10C2 = 45.

Event – the two pictures are the ones that Jim hasn’t already bought
Therefore, favourable outcomes = 7C2 = 21

Required probability = Favourable outcomes / Sample Space =  \frac{21 }{ 45}  =  \frac{7}{15} .

The backdoor approach to solving this question is to subtract the events where the 2 pictures selected to consist of pictures already bought by Jim.

There can be two such cases:
Both the pictures were the ones already bought by Jim, which can be done in 3C2 ways i.e. 3 ways.
One of the pictures was the one already bought by Jim and the other was one which wasn’t; this can be done in 3C1 * 7C1 ways i.e. 3*7 = 21 ways.

Therefore, required probability = 1 –  \frac{(3 + 21) }{ 45 } = 1 –  \frac{24 }{ 45}  =  \frac{21 }{ 45}  =  \frac{7 }{ 15} .

The correct answer option is D.

Hope that helps!
Aravind B T

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

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