In a race of 6 horses A, B, C, D, E and F, what is the probability of A finishing ahead of B and C both? There are no ties between any two horses.
A) 1/2
B) 1/3
C) 1/6
D) 1/12
E) 1/15
Source: www.GMATinsight.com
Answer: Option B
In a race of 6 horses A, B, C, D, E and F, what is the prob
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Within the race, the number of possible orderings for A, B and C = 3! = 6.GMATinsight wrote:In a race of 6 horses A, B, C, D, E and F, what is the probability of A finishing ahead of B and C both? There are no ties between any two horses.
A) 1/2
B) 1/3
C) 1/6
D) 1/12
E) 1/15
Since there are 2 orderings in which A appears before both B and C -- ABC and ACB -- we get:
P(A appears before both B and C) = 2/6 = 1/3.
The correct answer is B.
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Since we only care about A finishing ahead of B and C, we can ignore the other horses. If we only have three horses A, B and C, the ordering of the 3 horses can be:GMATinsight wrote:In a race of 6 horses A, B, C, D, E and F, what is the probability of A finishing ahead of B and C both? There are no ties between any two horses.
A) 1/2
B) 1/3
C) 1/6
D) 1/12
E) 1/15
Source: www.GMATinsight.com
Answer: Option B
ABC, ACB, BAC, BCA, CAB, CBA
We see that of the 6 orderings, only (the first) 2 have A ahead of B and C, therefore, the probability is 2/6 = 1/3.
Answer: B
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- Scott@TargetTestPrep
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Since we only care about A finishing ahead of B and C, we can ignore the other horses. If we only have three horses A, B and C, the ordering of the 3 horses can be:GMATinsight wrote:In a race of 6 horses A, B, C, D, E and F, what is the probability of A finishing ahead of B and C both? There are no ties between any two horses.
A) 1/2
B) 1/3
C) 1/6
D) 1/12
E) 1/15
Source: www.GMATinsight.com
Answer: Option B
ABC, ACB, BAC, BCA, CAB, CBA
We see that of the 6 orderings, only (the first) 2 have A ahead of B and C, therefore, the probability is 2/6 = 1/3.
Answer: B
Scott Woodbury-Stewart
Founder and CEO
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See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews