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If the probability that Mike can win a championship is 1/4, [#permalink]
 27 Jul 2007, 00:38
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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GMAT Club Legend
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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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Manager
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B.
You have to subtract the probability that Ben doesn't win.
(1/4)+(1/3)-(1/6)=5/12
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Manager
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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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Manager
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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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Director
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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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Director
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Re: Conceptual Probability Question [#permalink]
27 Jul 2007, 13:04
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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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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Director
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Anyone know the OA?
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Senior Manager
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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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Director
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i got B as well 5/12. What is the OA?
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Manager
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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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1/4 + 1/3 - 1/6 = 5/12
Ans B
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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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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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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  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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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 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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Director
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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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Senior Manager
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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.
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