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

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Re: Probability of Winning  [#permalink]

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New post 02 May 2010, 09:25
I dont agree with 7/12 as OA. I think it should be 25/72.

It is not mentioned that who starts the game.
What if the following happens:
1. MXX, XMX, XXM = 1/4 x 2/3 x 5/6 = 10/72
2. RXX, XRX, XXR = 1/3 x 3/4 x 5/6 = 15/72

All these 6 cases are valid. So, the prob = 25/72

Please let me know where I missed.

Skewed 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?

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Re: Probability of Winning  [#permalink]

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New post 09 May 2010, 09:40
chandrun wrote:
chandrun wrote:
7/12 and the reasoning seem very logical. However, looking at it from the other end, the probability of 'M' or 'R' winning is the same as that of 'B' losing. Since the probability of B winning is 1/6, the probability of B losing is 1-1/6 = 5/6. This seems pretty logical too :?:

Can somebody please explain why the above approach is wrong?


I think I realized my mistake. My previous argument would have made sense if the sum of the probabilities of M,R and B added to 1. Here, the probability of either one of M, R or B winning is just 3/4 i.e (1/4 + 1/3 + 1/6). So, the probability of none of the guys winning is 1/4.

The probability of B not winning is not the same as M or R winning since this discounts the case where all 3 of them lose.

Hence, the correct answer is (1/4 + 1/3) which amounts to 7/12. Alternatively, it can also be calculated as (3/4 - 1/6) which is also 7/12. 8-)


I agree with your reasoning; the sum of the probabilities of the 3 people winning is indeed less than 1.
Exactly for that reason, the correct answer should subtract the probability of Ben winning from the sum of the probabilities of Mike and Rob winning (since if both of them lose doesn't mean that Ben will win..)
So, imo the correct answer should be 1/4+1/3-1/6=5/12

Can someone plz confirm that?
What is the mistake I am making if any?
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Re: Probability of Winning  [#permalink]

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New post 24 Jun 2010, 12:27
1
Yoruichi wrote:
chandrun wrote:
chandrun wrote:
7/12 and the reasoning seem very logical. However, looking at it from the other end, the probability of 'M' or 'R' winning is the same as that of 'B' losing. Since the probability of B winning is 1/6, the probability of B losing is 1-1/6 = 5/6. This seems pretty logical too :?:

Can somebody please explain why the above approach is wrong?


I think I realized my mistake. My previous argument would have made sense if the sum of the probabilities of M,R and B added to 1. Here, the probability of either one of M, R or B winning is just 3/4 i.e (1/4 + 1/3 + 1/6). So, the probability of none of the guys winning is 1/4.

The probability of B not winning is not the same as M or R winning since this discounts the case where all 3 of them lose.

Hence, the correct answer is (1/4 + 1/3) which amounts to 7/12. Alternatively, it can also be calculated as (3/4 - 1/6) which is also 7/12. 8-)



I agree with your reasoning; the sum of the probabilities of the 3 people winning is indeed less than 1.
Exactly for that reason, the correct answer should subtract the probability of Ben winning from the sum of the probabilities of Mike and Rob winning (since if both of them lose doesn't mean that Ben will win..)
So, imo the correct answer should be 1/4+1/3-1/6=5/12

Can someone plz confirm that?
What is the mistake I am making if any?




HI we can solve this easily by the following steps!!!

P(a) = 1/4
p(b) = 1/3
p(c) = 1/6

the required probability is p( a U b @ c)

pls note [highlight]@ = intersection of[/highlight]

therefore P [ (aUb) @ c] = P(aUb) X P(c)
= [P(a) + P(b) - P(a)P(b) ] X P(c)
= [ 1/4 +1/3 - 1/12] X 5/6 = 5/12

Any further doubts pls let me know!!!
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Re: Probability of Winning  [#permalink]

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New post 25 Jun 2010, 12:27
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dixitraghav wrote:
good question


Hi!

Actually, I disagree. This is a bad question for a number of reasons.

First, it's ambiguous. The question says "can win a championship". There's no clarity as to whether all 3 participate in the same championship or if they participate in different ones; hence the confusion over dependent vs independent probability.

Second, there are no answer choices. Every question on the actual GMAT is, of course, accompanied by 5 choices. If you're not working with choices, you're not getting realistic GMAT prep.

Here are two ways the question could have been worded to resolve the ambiguity:

Quote:
Last year, Mike, Ben and Rob participated in a chess tournament that crowned 1 champion. If their respective probabilities of being crowned champion were 1/4, 1/3 and 1/6, then what's the probability that either Mike or Ben won?


Analysis: fairly low level probability question based on dependent events (dependent probability is quite rare on the GMAT).

Solution: \(\frac{1}{4} + \frac{1}{3} = \frac{7}{12}\).

Quote:
In 2007, Mike participated in a chess tournament in which he had a 1/4 chance of winning. In 2008, Ben participated in the same tournament and had a 1/3 chance to win. In 2009, Rob participated in that tournament and had a 1/6 chance to win. Assuming that all three events are independent, what's the probability that Mike and Ben won their tournaments but Rob didn't win his?


Analysis: medium level probability question based on independent events (independent probability is very common on the GMAT).

Solution: \(\frac{1}{4} * \frac{1}{3} * (1 - \frac{1}{6}) = (\frac{1}{4})(\frac{1}{3})(\frac{5}{6}) = \frac{5}{72}\)
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Re: Probability of Winning  [#permalink]

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New post 08 Aug 2010, 12:41
Sorry for opening this up again but I don't think these are dependent events, else total prob of winning will be 1 (but 1/4+1/3+1/6 = 3/4)

Hence they have to be either independent events or there has to be a 4th contestant as well.

In either case we can't "assume" that one will be the loser hence 25/72 looks correct

Thoughts??
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Re: Probability of Winning  [#permalink]

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New post 08 Aug 2010, 13:04
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gauravsaxena03 wrote:
Sorry for opening this up again but I don't think these are dependent events, else total prob of winning will be 1 (but 1/4+1/3+1/6 = 3/4)

Hence they have to be either independent events or there has to be a 4th contestant as well.

In either case we can't "assume" that one will be the loser hence 25/72 looks correct

Thoughts??


Hi, although the question is ambiguous for other reasons, the problem isn't the one that you've noted.

Nowhere does it say that the 3 people mentioned are the only participants in the tournament. In fact, since their probabilities don't add up to 100%, we know that either:

there are other participants; or
there doesn't have to be a winner.
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Re: Probability of Winning  [#permalink]

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New post 05 May 2011, 00:53
JuliaS wrote:
Hmm.. I got 35/72

Probability of Mike OR Rob winning is 1/4 + 1/3 = 7/12
Probability of Ben NOT winning is 6/6-1/6 = 5/6
Probability of Mike OR Rob winning and not Ben is 7/12*5/6 = 35/72

If this is not the right answer, can you pls. explain why this approach doesn't work? Thanks!


The third probability is wrong. You can do the multiplication only if 2 events are independent. P(A and B)=P(A).P(B) if and only if A and B are independent. This condition is not true in this problem because if (Mike or Rob) win then Ben CAN NOT win.

In fact, Probability( (Mike OR Rob) AND (Not BEN) ) = Probability(Mike OR Rob)*Probability(Not BEN, if Mike or Rob won). Probability(Not BEN, if Mike or Rob already won) is the probability that Ben do NOT win if Mike or Rob already won. Obviously, Probability(Not BEN, if Mike or Rob won) = 1.

Therefore Probability( (Mike OR Rob) AND (Not BEN) ) = Probability(Mike OR Rob)=7/12.

Correct me if I am wrong.
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Re: Probability of Winning  [#permalink]

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New post 05 May 2011, 01:15
25/72 is straight.
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Re: If the probability that Mike can win a championship is 1/4,  [#permalink]

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New post 03 Aug 2013, 03:00
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P(Either mike win or rob win) = P(mike win) + P(rob win) - P(mike and rob win)
P(Mike and rob win)=1/4 * 1/3 = 1/12

Thus P(either Mike or rob win)= 1/4 + 1/3 - 1/12 = 1/2

P(either Mike or rob win) and P(not Ben win) = 1/2 x 5/6 = 5/12.

I think this is correct.
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Re: If the probability that Mike can win a championship is 1/4,  [#permalink]

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