There are 5 competiros in a race (John, Mary and 3 other ones) in which there are no ties, what is the probability that Mary finish ahead of John?
A) 1/6
B) 1/3
C) 1/2
D) 2/3
E) 5/6
The OA is C.
Please, can any expert assist me with this PS question? I don't have it clear and I appreciate if any explain it for me. Thanks.
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There are 5 competitors in a race (John, mary and...
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DavidG@VeritasPrep
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Well, if we assume that the outcomes are all random, then there are only 2 possibilities: Mary finishes ahead of John or John finishes ahead of Mary, so there's a 1/2 chance that Mary finishes ahead of John. (Of course, if one of them is actually faster than the other, well that would create a bit of uncertainty.)AAPL wrote:There are 5 competiros in a race (John, Mary and 3 other ones) in which there are no ties, what is the probability that Mary finish ahead of John?
A) 1/6
B) 1/3
C) 1/2
D) 2/3
E) 5/6
The OA is C.
Please, can any expert assist me with this PS question? I don't have it clear and I appreciate if any explain it for me. Thanks.
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No uncertainty here: if I'm getting even odds, give me the faster runner for the max.DavidG@VeritasPrep wrote:(Of course, if one of them is actually faster than the other, well that would create a bit of uncertainty.)
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Scott@TargetTestPrep
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P(Mary finishes ahead of Jonn) = P(Mary finishes behind John)AAPL wrote:There are 5 competiros in a race (John, Mary and 3 other ones) in which there are no ties, what is the probability that Mary finish ahead of John?
A) 1/6
B) 1/3
C) 1/2
D) 2/3
E) 5/6
The OA is C.
Please, can any expert assist me with this PS question? I don't have it clear and I appreciate if any explain it for me. Thanks.
Since those are the only two ways the race can finish and since the number of outcomes where Mary finishes ahead of John is the same as the number of outcomes where John finishes ahead of Mary (if we have an outcome where Mary is ahead of John, just switch the positions of Mary and John and keep everyone else the same), the probability that Mary finishes ahead of John is 1/2.
Answer: C
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Hi All,
We're told that there are 5 competitors in a race (John, Mary and 3 others) in which there are no ties. We're asked for the probability that Mary finishes ahead of John. This is a great 'concept question', meaning that if you recognize the concepts involved, you do not have to do much math to answer the given question. As it stands, the number of competitors is IRRELEVENT - as long as there are at least 2 racers and there are no ties, the answer to this question will always be the SAME no matter how many racers there are.
Here's a simple example to prove it. What if it was just John and Mary? With 2 racers, the outcome would either be J first, M second or M first, J second.... meaning that there would be a 1/2 chance that Mary finishes ahead of John.
Adding a 3rd racer changes the total number of outcomes, but NOT the FRACTION in which Mary finishes ahead of John. There would be 6 outcomes (and I've labeled the third racer 'x')....
JMx
JxM
MJx
MxJ
xJM
xMJ
Mary finishes ahead of John 3 times out of the 6 possible outcomes. That's still 3/6 = 1/2 of the time.
With each additional racer, we will have additional possible outcomes, but the FRACTION in which Mary finishes ahead of John will be the same as before. It's always 1/2.
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
We're told that there are 5 competitors in a race (John, Mary and 3 others) in which there are no ties. We're asked for the probability that Mary finishes ahead of John. This is a great 'concept question', meaning that if you recognize the concepts involved, you do not have to do much math to answer the given question. As it stands, the number of competitors is IRRELEVENT - as long as there are at least 2 racers and there are no ties, the answer to this question will always be the SAME no matter how many racers there are.
Here's a simple example to prove it. What if it was just John and Mary? With 2 racers, the outcome would either be J first, M second or M first, J second.... meaning that there would be a 1/2 chance that Mary finishes ahead of John.
Adding a 3rd racer changes the total number of outcomes, but NOT the FRACTION in which Mary finishes ahead of John. There would be 6 outcomes (and I've labeled the third racer 'x')....
JMx
JxM
MJx
MxJ
xJM
xMJ
Mary finishes ahead of John 3 times out of the 6 possible outcomes. That's still 3/6 = 1/2 of the time.
With each additional racer, we will have additional possible outcomes, but the FRACTION in which Mary finishes ahead of John will be the same as before. It's always 1/2.
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
