Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 competitors in the triathlon, and medals are awarded for first, second, and third place, what is the probability that at least two of the triplets will win a medal?
(A) 3/14
(B) [spoiler]19/84[/spoiler]
(C) 11/42
(D) 15/28
(E) 3/4
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ace_gre
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Total number of people in the contest = A, B, C and 6 others.
Here order does not matter only people matter.
Total number of ways of selecting 3 out of 9 = 9C3 = 84
No. of ways of choosing 2 out of 3 triplets = 3C2
No. of ways of choosing 1 out of other 6 = 6C1
Ways of choosing 2 of 3 and one other winner = 3C2 * 6C1 = 18
No. of ways of choosing 3 out of 3 triplets = 3C3 = 1
Total ways of choosing >=2 triplets as winners = 18 +1 =19
Prob (>=2 triplets) = 19 / 84.
Here order does not matter only people matter.
Total number of ways of selecting 3 out of 9 = 9C3 = 84
No. of ways of choosing 2 out of 3 triplets = 3C2
No. of ways of choosing 1 out of other 6 = 6C1
Ways of choosing 2 of 3 and one other winner = 3C2 * 6C1 = 18
No. of ways of choosing 3 out of 3 triplets = 3C3 = 1
Total ways of choosing >=2 triplets as winners = 18 +1 =19
Prob (>=2 triplets) = 19 / 84.
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sanju09
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Good thing about this question is that each competitor can have at most one place for a medal.jagdeep wrote:Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 competitors in the triathlon, and medals are awarded for first, second, and third place, what is the probability that at least two of the triplets will win a medal?
(A) 3/14
(B) [spoiler]19/84[/spoiler]
(C) 11/42
(D) 15/28
(E) 3/4
Now, 3 places for a medal, 3 contenders out of 9 can be selected in 9C3 = 84 ways.
At least two of the triplets will be interpreted as either all 3 or any 2.
All 3 constitute a single combination, no matter who gets what, constituent probability = 1/84.
Any 2 of the triplet and one from the rest lot of 6 contenders can be had in 3C2*6C1 = 18 ways, constituent probability = 18/84.
Net required probability = 1/84 + 18/84 = [spoiler]19/84[/spoiler]. [spoiler]B[/spoiler]
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thephoenix
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another approach
probability of at least two can be calculated as 1-[p(0)+p(1)]
here p(0)=6c3/9c3=5/21
p(1)=3c1*6c2/9c3=15/28
p(0)+p(1)=5/21 + 15/28=65/84
P(>=2)=1-65/84=19/84
hth
probability of at least two can be calculated as 1-[p(0)+p(1)]
here p(0)=6c3/9c3=5/21
p(1)=3c1*6c2/9c3=15/28
p(0)+p(1)=5/21 + 15/28=65/84
P(>=2)=1-65/84=19/84
hth
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Brent@GMATPrepNow
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Let's using counting techniques.jagdeep wrote:Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 competitors in the triathlon, and medals are awarded for first, second, and third place, what is the probability that at least two of the triplets will win a medal?
(A) 3/14
(B) 19/84
(C) 11/42
(D) 15/28
(E) 3/4
P(at least 2 triplets win a medal) = (# of ways for at least 2 triplets win a medal)/(total # of ways to distribute the 3 medals)
As always, we'll begin with the denominator
total # of ways to distribute the 3 medals
There are 9 ways to award the 1st place medal
There are 8 ways to award the 2nd place medal
There are 7 ways to award the 3rd place medal
So, the total # of ways to distribute the 3 medals = (9)(8)(7)
# of ways for at least 2 triplets win a medal
We must consider two cases:
case a) all 3 triplets win medals
case b) 2 triplets win medals
case a) all 3 triplets win medals
We can arrange n unique objects in n! ways
So, we can arrange the 3 triplets (1st, 2nd, 3rd) in 3! ways (6 ways)
So, the total # of ways in which all 3 triplets win medals = 6
case b) 2 triplets win medals
First, we must determine the 3 athletes (2 triplets and 1 non-triplet) to win medals.
Select 2 of the 3 triplets. Using combinations, we can do this in 3C2 ways (3 ways)
Select 1 of the non-triplets. There are 6 non-triplets to choose from. So, we can do this in 6 ways
Now that we have our 3 athletes to win medals, we can arrange them (1st, 2nd, 3rd) in 3! ways (6 ways)
So, the total # of ways in which 2 triplets and 1 non-triplet win medals = (3)(6)(6) = 108
So, TOTAL # of ways for at least 2 triplets win a medal = 6 + 108 = 114
We get:
P(at least 2 triplets win a medal) = (# of ways for at least 2 triplets win a medal)/(total # of ways to distribute the 3 medals)
= 114/(9)(8)(7)
= 19/84
= B
Cheers,
Brent
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santhosh_katkurwar
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Thanks for the wonderful explaination Brent. Since the question used "atleast" in the question - I was wandering if we could this solve this using complement method in probability.
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Brent@GMATPrepNow
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thephoenix (above) uses the complement to solve the question.santhosh_katkurwar wrote:Thanks for the wonderful explaination Brent. Since the question used "atleast" in the question - I was wandering if we could this solve this using complement method in probability.
Cheers,
Brent
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Scott@TargetTestPrep
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The total number of ways to select 3 people from 9 is 9C3 = (9 x 8 x 7)/(3 x 2) = 3 x 4 x 7 = 84.jagdeep wrote:Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 competitors in the triathlon, and medals are awarded for first, second, and third place, what is the probability that at least two of the triplets will win a medal?
(A) 3/14
(B) [spoiler]19/84[/spoiler]
(C) 11/42
(D) 15/28
(E) 3/4
The number of ways 2 of the triplets are among the 3 awarded is 3C2 x 6C1 = 3 x 6 = 18.
The number of ways all 3 triplets are the 3 awarded is 3C3 x 6C0 = 1 x 1 = 1.
Therefore, the probability is (18 + 1)/84 = 19/84.
Alternate Solution:
The probability that one of the triplets is in first place and one of the triplets is in second place is 3/9 x 2/8 x 6/7 = 1/3 x 1/4 x 6/7 = 1/14.
Two of the top three positions can be occupied by the triplets in three ways (1st and 2nd, 1st and 3rd, 2nd and 3rd,) and each of these has the same probability, which we calculated above. Thus, the probability that exactly two of the triplets are in the top three positions is 3 x 1/14 = 3/14.
The probability that all three of the triplets occupy the top three positions is 3/9 x 2/8 x 1/7 = 1/84.
Thus, the probability that the triplets occupy at least two of the top three positions is 3/14 + 1/84 = 18/84 + 1/84 = 19/84.
Answer: B
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Could you please explain to me why order does not matter in this problem? I mean, there are actually 3, not 1, possible ways for the triplets to win a medal each. I don't understand why we are not taking order into account.Scott@TargetTestPrep wrote: ↑Thu Oct 17, 2019 7:23 pmThe total number of ways to select 3 people from 9 is 9C3 = (9 x 8 x 7)/(3 x 2) = 3 x 4 x 7 = 84.jagdeep wrote:Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 competitors in the triathlon, and medals are awarded for first, second, and third place, what is the probability that at least two of the triplets will win a medal?
(A) 3/14
(B) [spoiler]19/84[/spoiler]
(C) 11/42
(D) 15/28
(E) 3/4
The number of ways 2 of the triplets are among the 3 awarded is 3C2 x 6C1 = 3 x 6 = 18.
The number of ways all 3 triplets are the 3 awarded is 3C3 x 6C0 = 1 x 1 = 1.
Therefore, the probability is (18 + 1)/84 = 19/84.
Alternate Solution:
The probability that one of the triplets is in first place and one of the triplets is in second place is 3/9 x 2/8 x 6/7 = 1/3 x 1/4 x 6/7 = 1/14.
Two of the top three positions can be occupied by the triplets in three ways (1st and 2nd, 1st and 3rd, 2nd and 3rd,) and each of these has the same probability, which we calculated above. Thus, the probability that exactly two of the triplets are in the top three positions is 3 x 1/14 = 3/14.
The probability that all three of the triplets occupy the top three positions is 3/9 x 2/8 x 1/7 = 1/84.
Thus, the probability that the triplets occupy at least two of the top three positions is 3/14 + 1/84 = 18/84 + 1/84 = 19/84.
Answer: B
