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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 18 Jan 2008, 16:27
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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] New post 18 Jan 2008, 18:06
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?


P=(1/4 *2/3 * 5/6) + (1/3 * 3/4 * 5/6)=25/72
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Re: Probability of Winning [#permalink] New post 18 Jan 2008, 19:48
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!
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Re: Probability of Winning [#permalink] New post 18 Jan 2008, 20:09
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?


In this situation we are dealing with mutually exclusing events - there is only one person who wins the championship. If Mike or Rob win the championship it automatically means that Ben doesn't win the championship. The answer should be 1/4+1/3 = 7/12.
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Re: Probability of Winning [#permalink] New post 18 Jan 2008, 20:43
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?


1/3+1/4 = 7/12 1-1/6 = 5/6

Now 5/6*7/12 --> 35/72
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Re: Probability of Winning [#permalink] New post 18 Jan 2008, 22:20
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maratikus wrote:
In this situation we are dealing with mutually exclusing events - there is only one person who wins the championship. If Mike or Rob win the championship it automatically means that Ben doesn't win the championship. The answer should be 1/4+1/3 = 7/12.


Agree. The same logic.
good Q. +1
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Re: Probability of Winning [#permalink] New post 18 Jan 2008, 23:46
walker wrote:
maratikus wrote:
In this situation we are dealing with mutually exclusing events - there is only one person who wins the championship. If Mike or Rob win the championship it automatically means that Ben doesn't win the championship. The answer should be 1/4+1/3 = 7/12.


Agree. The same logic.
good Q. +1



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Re: Probability of Winning [#permalink] New post 19 Jan 2008, 05:51
can someone be clear with the concept & provide a correct solutionwithout confusing others? :x
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Re: Probability of Winning [#permalink] New post 19 Jan 2008, 08:43
walker wrote:
maratikus wrote:
In this situation we are dealing with mutually exclusing events - there is only one person who wins the championship. If Mike or Rob win the championship it automatically means that Ben doesn't win the championship. The answer should be 1/4+1/3 = 7/12.


Agree. The same logic.
good Q. +1



Well i thought so to. But apparently the OA is 5/12. The OE is "the probability that Mike or Rob will win is 1/4 + 1/3 or 7/12. The probability that NOT Ben will win is 1/6 and therefore, the total is 7/12 - 1/6 = 5/12". Seems like the OE is replete w/ mistakes.
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Re: Probability of Winning [#permalink] New post 19 Jan 2008, 12:29
Skewed wrote:
walker wrote:
maratikus wrote:
In this situation we are dealing with mutually exclusing events - there is only one person who wins the championship. If Mike or Rob win the championship it automatically means that Ben doesn't win the championship. The answer should be 1/4+1/3 = 7/12.


Agree. The same logic.
good Q. +1



Well i thought so to. But apparently the OA is 5/12. The OE is "the probability that Mike or Rob will win is 1/4 + 1/3 or 7/12. The probability that NOT Ben will win is 1/6 and therefore, the total is 7/12 - 1/6 = 5/12". Seems like the OE is replete w/ mistakes.


Something doesnt make sense here.

There is a problem in the GMATClub challenges that is almost identical to this problem. Except Bens probability not to win is 1/7.

The OA for that one is 1/2. Same process as I did in my above post, so maybe both the OA's are incorrect.
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Re: Probability of Winning [#permalink] New post 19 Jan 2008, 16:56
Skewed wrote:
This one is from Challenge 7.


Then the answer is 1/2 and ben's probabilty to lose is 1/7.
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Re: Probability of Winning [#permalink] New post 19 Jan 2008, 17:36
GMATBLACKBELT wrote:
Skewed wrote:
This one is from Challenge 7.


Then the answer is 1/2 and ben's probabilty to lose is 1/7.


Your right. However, the problem you speak of is a replacement of the one i posted. The answer to the previous question (the question in this post) is indeed 7/12 for the reasons Maratikus mentioned.
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Re: Probability of Winning [#permalink] New post 19 Jan 2008, 17:44
Skewed wrote:
walker wrote:
maratikus wrote:
In this situation we are dealing with mutually exclusing events - there is only one person who wins the championship. If Mike or Rob win the championship it automatically means that Ben doesn't win the championship. The answer should be 1/4+1/3 = 7/12.


Agree. The same logic.
good Q. +1



Well i thought so to. But apparently the OA is 5/12. The OE is "the probability that Mike or Rob will win is 1/4 + 1/3 or 7/12. The probability that NOT Ben will win is 1/6 and therefore, the total is 7/12 - 1/6 = 5/12". Seems like the OE is replete w/ mistakes.


In what results should I believe? 7/12 or 5/12. The later is not reliable, because I dont understand the above explained reasoning?

Please help!
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Re: Probability of Winning [#permalink] New post 19 Jan 2008, 22:31
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sondenso wrote:
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?


P=(1/4 *2/3 * 5/6) + (1/3 * 3/4 * 5/6)=25/72



I agree with you! That's how I solved it;

(Mikes wins*Rob looses*Ben Looses ) + ( Mike looses*Rob wins*Ben looses)

= 25/72.

Could someone tell what's wrong in this approach? this seems to be make sense.

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Re: Probability of Winning [#permalink] New post 19 Jan 2008, 23:53
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LM wrote:
sondenso wrote:
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?


P=(1/4 *2/3 * 5/6) + (1/3 * 3/4 * 5/6)=25/72



I agree with you! That's how I solved it;

(Mikes wins*Rob looses*Ben Looses ) + ( Mike looses*Rob wins*Ben looses)

= 25/72.

Could someone tell what's wrong in this approach? this seems to be make sense.



P=(1/4 *2/3 * 5/6) + (1/3 * 3/4 * 5/6)=25/72 is correct for independent events. For example, we have 3 different game. only one person take part in each game.

P=(1/4 *1 * 1) + (1/3 * 1 * 1)=5/12 is correct for dependent events. For example, we have 1 game. only one person can win. If Mike win the probabilities for Rob and Ben to not win are 1.
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Re: Probability of Winning [#permalink] New post 20 Jan 2008, 08:17
walker wrote:
LM wrote:
sondenso 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?


P=(1/4 *2/3 * 5/6) + (1/3 * 3/4 * 5/6)=25/72



I agree with you! That's how I solved it;

(Mikes wins*Rob looses*Ben Looses ) + ( Mike looses*Rob wins*Ben looses)

= 25/72.

Could someone tell what's wrong in this approach? this seems to be make sense.



P=(1/4 *2/3 * 5/6) + (1/3 * 3/4 * 5/6)=25/72 is correct for independent events. For example, we have 3 different game. only one person take part in each game.

P=(1/4 *1 * 1) + (1/3 * 1 * 1)=5/12 is correct for dependent events. For example, we have 1 game. only one person can win. If Mike win the probabilities for Rob and Ben to not win are 1.[/quote]
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You mean 1/4 + 1/3 = 7/12.
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Re: Probability of Winning [#permalink] New post 20 Jan 2008, 08:26
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Skewed wrote:
You mean 1/4 + 1/3 = 7/12.


Yes, thanks :)
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Re: Probability of Winning [#permalink] New post 27 Sep 2009, 05:02
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?

Soln: Probability that Mike or Bob win is
= Prob that Mike will win + prob that Bob will win
= 1/4 + 1/3
= 7/12
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Re: Probability of Winning [#permalink] New post 23 Mar 2010, 03:33
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?
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Re: Probability of Winning [#permalink] New post 23 Mar 2010, 03:54
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-)
Re: Probability of Winning   [#permalink] 23 Mar 2010, 03:54
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