A is one of the 6 horses enters for a race, and is to be

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A is one of the 6 horses enters for a race, and is to be [#permalink]

06 Jun 2004, 19:54

1. A is one of the 6 horses enters for a race, and is to be ridden by one of the 2 jockeys B & C. It is 2 to 1 that B rides A, in which case all the horses are equally likely to win; if C rides A, his chance is trebled: What are the odds against his winning?
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07 Jun 2004, 15:08

Not a fair gmat question. you should clearly explain, what does "odds of winning" mean? just as not everyone plays poker, not everyone bets on horses

that is the reason we dont have too many poker related questions on the gmat. its can be confusing to understand for some guys.

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07 Jun 2004, 15:26

Yes, this question is very ambiguous. What is meant by " It is 2 to 1 that B rides A"?
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07 Jun 2004, 15:37

"odds against winning" means "probability of loosing".

The phrase seems pretty reasonable to me but I dont have much experience on what is GMAT-formatted and what is not.

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07 Jun 2004, 17:29

GoalStanford wrote:

"odds against winning" means "probability of loosing".

The phrase seems pretty reasonable to me but I dont have much experience on what is GMAT-formatted and what is not.

goal, probability is a little different from odds. stick to the contents and the concepts in the OG.

Odds against is a ratio of the probability of A not occurring to the probability of A occurring.

Probability = favorable outcomes / total outcomes

ok?

I go by my experience with GMAT and the contents of the OG. when in doubt, see if the concept is there in the first few pages of the OG, where the concepts are explained.

Sincerely
Praet

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07 Jun 2004, 20:42

Okay,

it is 2 to 1 = 2/3
and
odds against winning - as far as I know - on the basis of which I have been doing probability and getting answers means probability of losing

one can rephrase the question on this basis and start working. I am sure with this rephrasing one can get a solution. It may not be a GMAT type, all the more it gives exposure to a tough problem. Anyway, one can ignore. But I am giving the solution here: -

There are two cases to consider:
(1) B rides, and (2) C rides.

(1) The probability that B rides is 2/3.
If B rides, the probabiity that A wins is 1/6. (Because, all the horses are equally likely to win)
Hence: P(B rides & A wins) = (2/3)(1/6) = 1/9

(2) The probability that C rides is 1/3.
If C rides, the probability that A wins is (3)(1/6) = 1/2.
Hence: P(C rides & A wins) = (1/3)(1/2) = 1/6.

Then: P(A wins) = 1/9 + 1/6 = 3/18 = 5/18 . . . and: P(A loses) = 13/18

The odds against his winning is 13 to 5
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Mayur

[#permalink] 07 Jun 2004, 20:42

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A is one of the 6 horses enters for a race, and is to be

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A is one of the 6 horses enters for a race, and is to be

A is one of the 6 horses enters for a race, and is to be

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