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by Uva@90 » Sun Aug 09, 2015 2:28 am
Jason, Aron and Dave along with six others students enter a competition. Each student has an equal chance of winning, and three prizes will be given, what is the probability that at least two of Josh, Aron and Dave will win prizes ?

A) 9/84
B) 17/84
C) 19/84
D) 23/84
E) 25/84

OA C
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by GMATGuruNY » Sun Aug 09, 2015 2:56 am
Uva@90 wrote:Jason, Aron and Dave along with six others students enter a competition. Each student has an equal chance of winning, and three prizes will be given, what is the probability that at least two of Josh, Aron and Dave will win prizes ?

A) 9/84
B) 17/84
C) 19/84
D) 23/84
E) 25/84
P(at least 2) = P(exactly 2) + P(all 3).

Case 1: Of Josh, Aron and Dave, exactly 2 are chosen
P(1st student is Josh, Aron or Dave) = 3/9. (Of the 9 students, 3 are Josh, Aron or Dave.)
P(2nd student is Josh, Aron or Dave) = 2/8. (Of the 8 remaining students, 2 are Josh, Aron or Dave.)
P(3rd student is NOT Josh, Aron or Dave) = 6/7. (Of the 7 remaining students, 6 are NOT Josh, Aron or Dave.)
To combine these probabilities, we MULTIPLY:
3/9 * 2/8 * 6/7 = 6/84.
Since the student who is not John, Aron or Dave could be selected 1st, 2nd or 3rd, we multiply by 3:
6/84 * 3 = 18/84.

Case 2: Josh, Aron and Dave are all chosen
P(1st student is Josh, Aron or Dave) = 3/9. (Of the 9 students, 3 are Josh, Aron or Dave.)
P(2nd student is Josh, Aron or Dave) = 2/8. (Of the 8 remaining students, 2 are Josh, Aron or Dave.)
P(3rd student is Josh, Aron or Dave) = 1/7. (Of the 7 remaining students, 1 is Josh, Aron or Dave.)
To combine these probabilities, we MULTIPLY:
3/9 * 2/8 * 1/7 = 1/84.

Thus:
P(at least 2) = 18/84 + 1/84 = 19/84.

The correct answer is C.
Last edited by GMATGuruNY on Sun Aug 09, 2015 10:37 am, edited 1 time in total.
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by Uva@90 » Sun Aug 09, 2015 7:23 am
GMATGuruNY wrote:
Uva@90 wrote:Jason, Aron and Dave along with six others students enter a competition. Each student has an equal chance of winning, and three prizes will be given, what is the probability that at least two of Josh, Aron and Dave will win prizes ?

A) 9/84
B) 17/84
C) 19/84
D) 23/84
E) 25/84
P(at least 2) = P(exactly 2) + P(all 3).

Case 1: Of Josh, Aron or Dave, exactly 2 are chosen
P(1st student is Josh, Aron or Dave) = 3/9. (Of the 9 students, 3 are Josh, Aron or Dave.)
P(2nd student is Josh, Aron or Dave) = 2/8. (Of the 8 remaining students, 2 are Josh, Aron or Dave.)
P(3rd student is NOT Josh, Aron or Dave) = 6/7. (Of the 7 remaining students, 6 are NOT Josh, Aron or Dave.)
To combine these probabilities, we MULTIPLY:
3/9 * 2/8 * 6/7 = 6/84.
Since the student who is not John, Aron or Dave could be selected 1st, 2nd or 3rd, we multiply by 3:
6/84 * 3 = 18/84.

Case 2: Josh, Aron and Dave are all chosen
P(1st student is Josh, Aron or Dave) = 3/9. (Of the 9 students, 3 are Josh, Aron or Dave.)
P(2nd student is Josh, Aron or Dave) = 2/8. (Of the 8 remaining students, 2 are Josh, Aron or Dave.)
P(3rd student is Josh, Aron or Dave) = 1/7. (Of the 7 remaining students, 1 is Josh, Aron or Dave.)
To combine these probabilities, we MULTIPLY:
3/9 * 2/8 * 1/7 = 1/84.

Thus:
P(at least 2) = 18/84 + 1/84 = 19/84.

The correct answer is C.
Thanks a Lot Mitch.

I should add this to my Dictionary.
P(at least 2) = P(exactly 2) + P(all 3).
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