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!
A set of pictures of butterflies contains 10 pictures.
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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>
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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P(1st picture selected has not already been purchased) = 7/10. (Of the 10 pictures, 3 have already been purchased, so 7 have not.)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(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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The number of ways to select 2 pictures from 10 is 10C2 = 10!/(2! x 8!) = (10 x 9)/2 = 45.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 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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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
________________________________________
Shahrukh Moin Khan
QA Mentor at Breakspace
https://www.mbabreakspace.com
PGDM: IIM Calcutta
B.Tech IIT Roorkee
No. of ways= 7C2= 21
Total number of ways to select 2 out of 10 = 10C2= 45
Probability= 21/45= 7/15
________________________________________
Shahrukh Moin Khan
QA Mentor at Breakspace
https://www.mbabreakspace.com
PGDM: IIM Calcutta
B.Tech IIT Roorkee