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A set of pictures of butterflies contains 10 pictures.

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A set of pictures of butterflies contains 10 pictures.

by AAPL » Wed Jun 27, 2018 5:36 am

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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 no those that were already bought by Jim?

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

The OA is D.

Jim selects 3 pictures. So the 2 pictures are to be selected from the remaining 7.

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

2 pictures to be selected from 7 (i.e. pictures no chosen by Jim) = 7C2 = 21.

The required probability = 21/45 = 7/15.

Has anyone another strategic approach to solve this PS question? Regards!
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by Vincen » Thu Jun 28, 2018 2:04 am
Hello AAPL.

I will show you another approach but this one is a little bit longer.

We know the following $$P\left(both\ pictures\ are\ no\ those\right)+P\left(at\ least\ one\ picture\ is\ one\ of\ those\right)=1.$$ As you said, the number of ways that 2 pictures can be selected from 10 is $$10\ C\ 2=\frac{10!}{8!\cdot2!}=45.$$ Now, $$P\left(at\ least\ one\ picture\ is\ one\ of\ those\right)$$ can be splited in two parts:

i) Pick only one picture from the selected ones:

In this case, there are 21 different ways since $$\left(select\ one\ of\ the\ three\right)\cdot\left(select\ one\ from\ the\ other\ 7\right)=\left(3\ C\ 1\right)\cdot\left(7\ C\ 1\right)=3\cdot7=21.$$

ii) Pick both 2 pictures from the 3 selected ones:

In this case, there are 3 different ways since $$3\ C\ 2\ =\frac{3!}{1!\cdot2!}=3.$$

Hence we have: $$P\left(both\ pictures\ are\ no\ those\right)+P\left(at\ least\ one\ picture\ is\ one\ of\ those\right)=1$$ $$P\left(both\right)+\left(\frac{3}{45}+\frac{21}{45}\right)=1$$ $$P\left(both\right)+\left(\frac{1}{15}+\frac{7}{15}\right)=1$$ $$P\left(both\right)+\left(\frac{8}{15}\right)=1$$ $$P\left(both\right)=1-\frac{8}{15}$$ $$P\left(both\right)=\frac{7}{15}.$$ Therefore, the correct answer is the option D.

I hope it helps you. <i class="em em-smiley"></i>
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by GMATGuruNY » Thu Jun 28, 2018 2:15 am
AAPL 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 no those that were already bought by Jim?

A. 14/15
B. 3/5
C. 6/13
D. 7/15
E. 7/10
P(1st picture selected has not already been purchased) = 7/10. (Of the 10 pictures, 3 have already been purchased, so 7 have not.)
P(2nd picture selected has not already been purchased) = 6/9. (Of the 9 remaining pictures, 3 have already been purchased, so 6 have not.)
To combine these probabilities, we multiply:
7/10 * 6/9 = 7/15.

The correct answer is D.
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by Jeff@TargetTestPrep » Mon Jul 02, 2018 9:29 am
AAPL 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 no 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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by Shahrukh_mbabreakspace » Tue Jul 03, 2018 1:49 am
We need to select 2 out 7 remaining
No. of ways= 7C2= 21

Total number of ways to select 2 out of 10 = 10C2= 45

Probability= 21/45= 7/15

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