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Odds and Probability

Expert replies
by dtweah » Sat Aug 14, 2010 3:03 pm
In an essay competition the odds against competitors A, B, C, D are 2;1, 3;1, 4;1 and 5;1. Find the probability that one of them wins the competition?

A. 11/ 120

B. 114/120

C. 127/150

D. 2/17

E. 135/157

[spoiler]https://www.greguide.com/Free-GRE-Practi ... est-5.html[/spoiler]
Last edited by dtweah on Sat Aug 14, 2010 5:12 pm, edited 1 time in total.
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Source: — Problem Solving |

by Stuart@KaplanGMAT » Sat Aug 14, 2010 4:57 pm
dtweah wrote:In an essay competition the odds against competitors A, B, C, D are 2;1, 3;1, 4;1 and 5;1. Find the probability that one of them wins the competition?

A. 11/ 120

B. 114/120

C. 127/140

D. 2/17

E. 135/157

[spoiler]https://www.greguide.com/Free-GRE-Practi ... est-5.html[/spoiler]
Hi,

if the probability against you winning is 2:1, then you have a 1/3 chance of winning. Applying the same rule to the remaining contestants, they have 1/3, 1/4, 1/5 and 1/6 chances to win.

Since we want the probability of A or B or C or D winning, we ADD the individual probabilities.

So:

1/3 + 1/4 + 1/5 + 1/6

getting a common denominator of 120:

40/120 + 30/120 + 24/120 + 20/120

= 114/120... choose B.

Note, however, that on the real test the answer would be 19/20, since answers are always reduced as much as possible. Another example of a poorly written question from this source (the question itself has ambiguities as well).
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by dtweah » Sat Aug 14, 2010 7:29 pm
Stuart Kovinsky wrote:
dtweah wrote:In an essay competition the odds against competitors A, B, C, D are 2;1, 3;1, 4;1 and 5;1. Find the probability that one of them wins the competition?

A. 11/ 120

B. 114/120

C. 127/140

D. 2/17

E. 135/157

[spoiler]https://www.greguide.com/Free-GRE-Practi ... est-5.html[/spoiler]
Hi,

if the probability against you winning is 2:1, then you have a 1/3 chance of winning. Applying the same rule to the remaining contestants, they have 1/3, 1/4, 1/5 and 1/6 chances to win.

Since we want the probability of A or B or C or D winning, we ADD the individual probabilities.

So:

1/3 + 1/4 + 1/5 + 1/6

getting a common denominator of 120:

40/120 + 30/120 + 24/120 + 20/120

= 114/120... choose B.

Note, however, that on the real test the answer would be 19/20, since answers are always reduced as much as possible. Another example of a poorly written question from this source (the question itself has ambiguities as well).

OA different Stuart.
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by Stuart@KaplanGMAT » Sat Aug 14, 2010 7:40 pm
dtweah wrote:
OA different Stuart.
Then I'd love to see the OA and explanation - I don't see an error in my approach.
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by Ian Stewart » Sat Aug 14, 2010 11:58 pm
dtweah wrote:
OA different Stuart.
If the OA is different, then find better practice material. Stuart's work is all perfectly correct (as usual!).
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com
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by mainhoon » Sun Aug 15, 2010 7:35 pm
Ian Stewart wrote:
dtweah wrote:
OA different Stuart.
If the OA is different, then find better practice material. Stuart's work is all perfectly correct (as usual!).
Indeed, the source of material is critical. Especially in the Verbal area! One tends to get more confused than get real answers.. Ian, I have posted a combination question, please take a look and comment.. Thanks
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