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If the probability that Mike can win a championship is 1/4,

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If the probability that Mike can win a championship is 1/4, [#permalink] New post 26 Jul 2007, 23:38
If the probability that Mike can win a championship is 1/4, and that of Rob winning is 1/3 and that of Ben winning is 1/6, what is the probability that Mike or Rob win the Championship but not Ben ?

1) 1/6
2) 5/12
3) 7/12
4) 5/6
5) 1
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 [#permalink] New post 27 Jul 2007, 01:05
Only one person can be the champion, and the question says either Rob or Ben win but not mike (i'm not reading wrong, i hope)

P(rob win) + P(mike win) = 1/4 + 1/3 = 7/12
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 [#permalink] New post 27 Jul 2007, 05:05
B.

You have to subtract the probability that Ben doesn't win.

(1/4)+(1/3)-(1/6)=5/12
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 [#permalink] New post 27 Jul 2007, 09:56
Hi briks123

The way i solved it was

Mike will Win + Rob Will Win + Ben will not win =

1/4 + 1/3 + 5/6 = 17/12

But am going wrong somewhere obviously.. Could you help me out.. :)
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 [#permalink] New post 27 Jul 2007, 10:03
I got 5/12 using same method as briks123.

Ajay, I think what went wrong with your method is when you calculate P(Ben will not win), it also include the chances of Mike or Rob winning.
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 [#permalink] New post 27 Jul 2007, 11:38
I am with Briks and Ankita.

P(M)+P(R)-P(M*R)=1/2
1/2*(1-P(B))=5/12
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Re: Conceptual Probability Question [#permalink] New post 27 Jul 2007, 12:04
ajay_gmat wrote:
If the probability that Mike can win a championship is 1/4, and that of Rob winning is 1/3 and that of Ben winning is 1/6, what is the probability that Mike or Rob win the Championship but not Ben ?

1) 1/6
2) 5/12
3) 7/12
4) 5/6
5) 1


imo, it is

(prob that mike or rob win) (prob that ben won't win) = (1/4 + 1/3) (1 - 1/6) = 35/72
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 [#permalink] New post 27 Jul 2007, 17:56
I am with Himalaya...
Since prob of winning either of three of them does not sum to 1 so there are other persons as well.

So it is will:
(prob of mike or rob winning) * prob of Ben not winnning

(1/3 +1/4) * (1-1/6)= 7/12 * 5/6 =35/72
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 [#permalink] New post 27 Jul 2007, 18:04
Anyone know the OA?
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 [#permalink] New post 30 Jul 2007, 15:53
ajay_gmat wrote:
Hi briks123

The way i solved it was

Mike will Win + Rob Will Win + Ben will not win =

1/4 + 1/3 + 5/6 = 17/12

But am going wrong somewhere obviously.. Could you help me out.. :)


they are all exclusive events
(i.e. both people can't win) therefore they are not independant events

Pr A or B = Pr( A) + Pr (B) - Pr( A ANDB)

1/4+1/3=7/12

it might be conceptually simpler to construct two complementary events that sum to 1, PR Ben will win + PR Ben won't win - but that doesn't work in this situation because like others stated we are missing another person.
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 [#permalink] New post 31 Jul 2007, 22:38
i got B as well 5/12. What is the OA?
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 [#permalink] New post 01 Aug 2007, 03:59
ajay_gmat wrote:
OA is 5/12


To be honest, I had no idea that you would subtract the other guys probability until I saw a similar post on GMAT club.

So "either A or B but not C" implies you add A's and B's probabilities and subtract C's probability.
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 [#permalink] New post 01 Aug 2007, 17:10
1/4 + 1/3 - 1/6 = 5/12

Ans B
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 [#permalink] New post 01 Aug 2007, 17:30
cruiser wrote:
1/4 + 1/3 - 1/6 = 5/12

Ans B


why are we subtracting Ben's probability of winning?

P(AorB) = P(A) + P(B) - (PAandB)

there can be only one winner so P(AandB)=0

saying P A or B but not C has no meaning because if A or B wins, C can't win.


we are missing another contestant, 'D', with a 3/12 chance of winning... C+D = 5/12.

just seems weird to me.
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 [#permalink] New post 02 Aug 2007, 08:36
anonymousegmat wrote:
cruiser wrote:
1/4 + 1/3 - 1/6 = 5/12

Ans B


why are we subtracting Ben's probability of winning?

P(AorB) = P(A) + P(B) - (PAandB)

there can be only one winner so P(AandB)=0

saying P A or B but not C has no meaning because if A or B wins, C can't win.


we are missing another contestant, 'D', with a 3/12 chance of winning... C+D = 5/12.

just seems weird to me.


Yeah I don't really get it... but that's the math rule or something
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 [#permalink] New post 03 Aug 2007, 09:56
briks123 wrote:
anonymousegmat wrote:
cruiser wrote:
1/4 + 1/3 - 1/6 = 5/12

Ans B


why are we subtracting Ben's probability of winning?

P(AorB) = P(A) + P(B) - (PAandB)

there can be only one winner so P(AandB)=0

saying P A or B but not C has no meaning because if A or B wins, C can't win.


we are missing another contestant, 'D', with a 3/12 chance of winning... C+D = 5/12.

just seems weird to me.


Yeah I don't really get it... but that's the math rule or something


:evil: i hate rules that i don't understand. i will get any problem of this ilk wrong on the gmat, because i don't understand the logic behind it.

this is GMAT prep at its worst. employing a rule/concept you don't understand because you are told it is right.
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 [#permalink] New post 03 Aug 2007, 13:06
anonymousegmat wrote:
briks123 wrote:
anonymousegmat wrote:
cruiser wrote:
1/4 + 1/3 - 1/6 = 5/12

Ans B


why are we subtracting Ben's probability of winning?

P(AorB) = P(A) + P(B) - (PAandB)

there can be only one winner so P(AandB)=0

saying P A or B but not C has no meaning because if A or B wins, C can't win.


we are missing another contestant, 'D', with a 3/12 chance of winning... C+D = 5/12.

just seems weird to me.


Yeah I don't really get it... but that's the math rule or something


:evil: i hate rules that i don't understand. i will get any problem of this ilk wrong on the gmat, because i don't understand the logic behind it.

this is GMAT prep at its worst. employing a rule/concept you don't understand because you are told it is right.

Just think about it this way:
The total probability at least ONE of them wins:
P(Mike Wins) + P(Rob Wins) + P(Ben Wins) = 9/12

If they told you that you could not count on Ben winning, so now what are the chances of total win between Mike and Rob with Ben's loss?
-- Here you're subtracting the possibility of Ben winning all together from their combined probability. So:
P(Mike) + P(Rob) - P(Ben) = 7/12
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 [#permalink] New post 03 Aug 2007, 17:10
Probability of Ben not winning is either Mike or Rob winning the Championship.

This implies: 1 – 1/6 = 5/6

Answer Choice: 4
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 [#permalink] New post 03 Aug 2007, 19:34
hanumayamma wrote:
Probability of Ben not winning is either Mike or Rob winning the Championship.

This implies: 1 – 1/6 = 5/6

Answer Choice: 4


that would be true if the probabilties summed to 1, but they don't. there are other contestants.
  [#permalink] 03 Aug 2007, 19:34
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