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Difficulty:
75%
(hard)
Question Stats:
36% (02:47) correct
64%
(03:11)
wrong
based on 14
sessions
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Let
P = {0.375,1.25,2, 3,4,5,5.5}
If three distinct numbers are selected at random from \(P\), what is the probability that their product is an integer?
A. \(\frac{2}{7}\)
B. \(\frac{13}{35}\)
C. \(\frac{2}{5}\)
D. \(\frac{16}{35}\)
E. \(\frac{18}{35}\)
P = {0.375,1.25,2, 3,4,5,5.5}
If three distinct numbers are selected at random from \(P\), what is the probability that their product is an integer?
A. \(\frac{2}{7}\)
B. \(\frac{13}{35}\)
C. \(\frac{2}{5}\)
D. \(\frac{16}{35}\)
E. \(\frac{18}{35}\)
![P = {0.375, 1,25, 2, 3, 4, 5.5}[/m]. If three](/forum/styles/gmatclub_light/imageset/icon_post_target_unread.svg "P = {0.375, 1,25, 2, 3, 4, 5.5}[/m]. If three"){kind=icon}
28 Apr 2026, 20:23
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Solution for now edited question
P is as given in the title {0.375,1.25,2, 3, 4,5,5.5}
Write them as fraction to make it easier.
{0.375=3*0.125=3*1/8,1.25=5/4,2, 3, 4,5,5.5=11/2}
The different ways three distinct number will give you an integer as their product.
1) All three integers...
4C3 or 4 as there are exactly four integers.
2) One fraction and other two integers
a) 3/8.. Only 2*4 possible, so 1 way
b) 11/2.. one has to be a multiple of 2 and second any other integer.. 2 and any of remaining 3 so 3C1 ways. If you take 4 and second any from remaining 2(excluding 2 as 2*4 already catered above) so 2C1 ways. Thus 3+2 or 5 ways
c) 5/4..one number is 4 and other can be any other integer, so 3C1 or 3 ways
Total 1+5+3 or 9 ways
3) Two fractions and one integer..
No way possible as you would require the third to be 8 or its multiple.
Total possibilities- 4+9 ie 13 ways.
Overall ways - 7C3 or 35 ways
Probability = 13/35
_____________________________________________
I am leaving solution for other two cases from unedited questions.
If set P is as given above {0.375,1.25,2, 3, 4,5.5}
Write them as fraction to make it easier.
{0.375=3*0.125=3*1/8,1.25=5/4,2, 3, 4,5.5=11/2}
The different ways three distict number will give you an integer as their product.
1) All three integers...
3C3 or 1 as there are exactly three integers.
2) One fraction and other two integers
a) 3/8.. Only 2*4 possible, so 1 way
b) 5/4..one number is 4 and other can be any other integer, so 2C1 or 2 ways
c) 11/2.. one has to be a multiple of 2 and second any other integer.. 2*3, 2*4 or 3*4, so 3 ways.
Total 6 ways
3) Two fractions and one integer..
No way possible as you would require the third to be 8 or its multiple.
Total possibilities- 1+6 ie 7 ways.
Overall ways - 6C3 or 20 ways
Probability = 7/20
But say P is as given in the title {0.375,1,25,2, 3, 4,5.5}
Write them as fraction to make it easier.
{0.375=3*0.125=3*1/8,1, 25,2, 3, 4,5.5=11/2}
The different ways three distinct number will give you an integer as their product.
1) All three integers...
5C3 or 10 as there are exactly five integers.
2) One fraction and other two integers
a) 3/8.. Only 2*4 possible, so 1 way
b) 11/2.. one has to be a multiple of 2 and second any other integer.. 2 and any of remaining 4 so 4C1 ways. If you take 4 and second any from remaining 3(excluding 2 as 2*4 already catered above) so 3C1 ways. Thus 4+3 or 7 ways
Total 1+7 or 8 ways
3) Two fractions and one integer..
No way possible as you would require the third to be 8 or its multiple.
Total possibilities- 10+8 ie 18 ways.
Overall ways - 7C3 or 35 ways
Probability = 18/35
In both cases the OA is wrong. If the set is correct as in the title, the answer is E.
