# Probability of winning competition given odds

• Sep 30th 2009, 05:30 PM
devins
Probability of winning competition given odds
In an essay competition the odds against competitors A, B, C, and D are 2:1, 3:1, 4:1, and 5:1 respectively. Find the probability that one of them wins the competition.

The answer is 127/150 but I cannot find that answer either online or by myself. I know that the odds against them translates to probabilities of 2/3, 3/4, 4/5, and 5/6 respectively but for starters those numbers do not add up to one. Subtracting one from those numbers and adding them sums to 0.95, which is helpful and intuitive but not the answer. I tried finding the probabilities of each one winning and the others losing and adding those but that was not correct either. Any help is greatly appreciated because this kind of stuff drives me crazy. Here is the URL of the question (first one of the practice GRE test):

GRE Practice Test - Problem Solving GRE Practice Test 5
• Oct 3rd 2009, 12:03 PM
awkward
Quote:

Originally Posted by devins
In an essay competition the odds against competitors A, B, C, and D are 2:1, 3:1, 4:1, and 5:1 respectively. Find the probability that one of them wins the competition.

The answer is 127/150 but I cannot find that answer either online or by myself. I know that the odds against them translates to probabilities of 2/3, 3/4, 4/5, and 5/6 respectively but for starters those numbers do not add up to one. Subtracting one from those numbers and adding them sums to 0.95, which is helpful and intuitive but not the answer. I tried finding the probabilities of each one winning and the others losing and adding those but that was not correct either. Any help is greatly appreciated because this kind of stuff drives me crazy. Here is the URL of the question (first one of the practice GRE test):

GRE Practice Test - Problem Solving GRE Practice Test 5

Hi Devins,

I'm going to stick my neck out and say I believe the answer given on the site is wrong.

As you said, the probabilities that A-D do not win the contest are 2/3, 3/4, 4/5, and 5/6 respectively.

So the probability of winning is 1/3 for A, 1/4 for B, 1/5 for C and 1/6 for D.

Since only one person can win, these are mutually exclusive events.

So the probability that A, B, C, or D wins is 1/3 + 1/4 + 1/5 + 1/6 = 19/20.
• Oct 3rd 2009, 10:05 PM
mr fantastic
Quote:

Originally Posted by awkward
Hi Devins,

I'm going to stick my neck out and say I believe the answer given on the site is wrong.

As you said, the probabilities that A-D do not win the contest are 2/3, 3/4, 4/5, and 5/6 respectively.

So the probability of winning is 1/3 for A, 1/4 for B, 1/5 for C and 1/6 for D.

Since only one person can win, these are mutually exclusive events.

So the probability that A, B, C, or D wins is 1/3 + 1/4 + 1/5 + 1/6 = 19/20.

I was going to say the same thing but decided to think about it some more (I've already made three careless mistakes in the last 4 days and didn't want to add to the list) and then never got around to replying.
• Oct 5th 2009, 01:32 PM
devins
Thanks!
Hey thanks everyone! You have saved me from many headaches. Can't believe that GRE website would post an incorrect answer!! It ought to be illegal. Anyhow, thanks for the help!