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If Ben were to lose the championship, Mike would be the winner with a [#permalink]

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02 Oct 2010, 05:57

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If Ben were to lose the championship, Mike would be the winner with a probability of \(\frac{1}{4}\), and Rob - \(\frac{1}{3}\) . If the probability of Ben being the winner is \(\frac{1}{7}\), what is the probability that either Mike or Rob will win the championship? Assume that there can be only one winner.

Re: If Ben were to lose the championship, Mike would be the winner with a [#permalink]

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02 Oct 2010, 07:12

Barkatis wrote:

The OA : \((1-\frac{1}{7}) * \frac{7}{12} = \frac{6}{7} * \frac{7}{12} = \frac{1}{2}\) . (7/12 = 1/3 + 1/4)

Can Some one explain me what is the problem with this reasoning : P=(1/4 *2/3 * 6/7) + (1/3 * 3/4 * 6/7)= 5/14

Thanks

I think what you have done is :

Probability = P(Ben loses) * P(Mike wins given Ben loses) * (1 - P(Rob wins givne Ben loses)) + P(Ben loses) * P(Rob wins given Ben loses) * (1 - P(Mike wins givne Ben loses))

The problem is the terms I have marked in red. You are already given the probability that Mike wins once Ben has lost, 1/4, you do not need to multiply this with (1-2/3). Similarly in the second term

The correct answer would be : P(Ben loses) * P(Mike wins given Ben loses) + P(Ben loses) * P(Rob wins given Ben loses) OR (6/7) * (1/3) + (6/7) * (1/4) = (1/2)
_________________

Re: If Ben were to lose the championship, Mike would be the winner with a [#permalink]

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02 Oct 2010, 12:11

Thanks for answering.

I got that part but my question is how do we know that the probability given is that Mike wins ONCE BEN HAS LOST.

What is the difference between that problem and this one for example Xavier, Yvonne, and Zelda each try independently to solve a problem. If their individual probabilities for success are 1/4 ,1/2 , and 5/8 , respectively, what is the probability that Xavier and Yvonne, but not Zelda, will solve the problem ?

Where the solution is 3/8*1/2*1/4 = 3/64 But if we assume that the probability given are those of success of one person while the two others are loosing it would be more : 1/4*1/2 . Right ?

Re: If Ben were to lose the championship, Mike would be the winner with a [#permalink]

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02 Oct 2010, 17:26

Barkatis wrote:

Thanks for answering.

I got that part but my question is how do we know that the probability given is that Mike wins ONCE BEN HAS LOST.

What is the difference between that problem and this one for example Xavier, Yvonne, and Zelda each try independently to solve a problem. If their individual probabilities for success are 1/4 ,1/2 , and 5/8 , respectively, what is the probability that Xavier and Yvonne, but not Zelda, will solve the problem ?

Where the solution is 3/8*1/2*1/4 = 3/64 But if we assume that the probability given are those of success of one person while the two others are loosing it would be more : 1/4*1/2 . Right ?

I hope you got my problem better now.

You have to be careful about reading the wording. In the example you are giving, there are 3 people solving a question. None, one or more could solve it correctly. In the question at hand, there are 3 players trying to win a tournament. Either none of them or atmost one of them can win, and simultaneous winning is not possible.

Now whether we have the probability of Mike winning given Ben has lost or we have absolute probability of Mike winning is just how the question is worded.

If Ben were to lose the championship,Mike would be the winner with a probability of \frac{1}{4}

Re: If Ben were to lose the championship, Mike would be the winner with a [#permalink]

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21 Mar 2011, 07:08

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asmit123 wrote:

If Ben were to lose the championship,MIke would be the winner with a probability of 1/4, and Rob - 1/3. If probab of Ben winning is 1/7, what is the probab that either Mike or Rob will win the championship(there can be only one winner) ?

1.1/12 2.1/7 3.1/2 4.7/12 5.6/7

Probability Ben will lose =1-1/7 =>6/7 Probability Ben will lose and Mike will win => 1/4*6/7 Probability Ben will lose and Rob will win=>1/3*6/7

Probability Either Mike or Rob will win=> 6/7(1/4+1/3) 6/7 *7/12=> 1/2

This question is based on "conditional" probability, which is an exceptionally rare subject on the GMAT (you probably will NOT see it at all on Test Day - if you do, then it would likely be in a DS question). Unless you're already scoring in the Q49+ range, you be spending your time on other subjects.

If Ben were to lose the championship, Mike would be the winner with a probability of \(\frac{1}{4}\), and Rob - \(\frac{1}{3}\) . If the probability of Ben being the winner is \(\frac{1}{7}\), what is the probability that either Mike or Rob will win the championship? Assume that there can be only one winner.

A. \(\frac{1}{12}\) B. \(\frac{1}{7}\) C. \(\frac{1}{2}\) D. \(\frac{7}{12}\) E. \(\frac{6}{7}\)

M07-12

This is a conditional probability question. We need the probability that either Mike or Rob will win the championship. So Ben must lose: the probability of Ben losing is \(1-\frac{1}{7}=\frac{6}{7}\).

Now out of these \(\frac{6}{7}\) cases the probability of Mike winning is \(\frac{1}{4}\) and the probability of Rob winning is \(\frac{1}{3}\). So \(P=\frac{6}{7}(\frac{1}{4}+\frac{1}{3})=\frac{1}{2}\).

Or consider the following:

Take 84 championships/cases (I chose 84 as it's a LCM of 3, 4, and 7).

Now, out of these 84 cases Ben will lose in \(\frac{6}{7}*84=72\). Mike would be the winner in \(72*\frac{1}{4}=18\) (1/4 th of the cases when Ben loses) and Rob would be the winner in \(72*\frac{1}{3}=24\). Therefore \(P=\frac{18+24}{84}=\frac{1}{2}\).

Re: If Ben were to lose the championship, Mike would be the winner with a [#permalink]

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15 Jul 2015, 22:26

Hi All,

I am confused..i thought the probability that Ben will lose = P of either Mike r Rob will win? or we are assuming that there are still others in the championship? so my earlier assumption is not valid?

Re: If Ben were to lose the championship, Mike would be the winner with a [#permalink]

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15 Jul 2015, 23:00

katzzzz wrote:

Hi All,

I am confused..i thought the probability that Ben will lose = P of either Mike r Rob will win? or we are assuming that there are still others in the championship? so my earlier assumption is not valid?

Thanks in advance

Hi,

I don't quite understand your doubt, see if this helps...

Probability that Ben wins = 1/7 (given); so the 'counter-event' would be Ben losing = 1 - 1/7 = 6/7.

Now, they have asked to find the probability that EITHER Rob or Mike wins, and there are 2 conditions imposed: (1) there can only be one winner (2) (stemming from the 1st condition) Ben loses (because we have to find only the probability of Mike and Rob, which means Ben loses as there can be only one winner).

Rob winning the game given that Ben had lost = 1/3 * 6/7 Mike winning given that Ben had lost = 1/4 * 6/7

Adding those two we get the answer.
_________________

I used to think the brain was the most important organ. Then I thought, look what’s telling me that.

Re: If Ben were to lose the championship, Mike would be the winner with a [#permalink]

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16 Jul 2015, 02:00

Hi,

I mean..why can't we choose 6/7 as the answer. Probability of either one of them win = probability of mike lost the game? I not sure where did get it wrong.. Thanks

I mean..why can't we choose 6/7 as the answer. Probability of either one of them win = probability of mike lost the game? I not sure where did get it wrong.. Thanks

hi, the probability given for other two to be winner shows that there are other in the race.. otherwise the total probability should have been 1 for all three winning ..
_________________

Re: If Ben were to lose the championship, Mike would be the winner with a [#permalink]

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16 Jul 2015, 03:13

katzzzz wrote:

Hi,

I mean..why can't we choose 6/7 as the answer. Probability of either one of them win = probability of mike lost the game? I not sure where did get it wrong.. Thanks

The most simple and direct answer to that question is the events are independent. I hope you get it now?
_________________

I used to think the brain was the most important organ. Then I thought, look what’s telling me that.

Re: If Ben were to lose the championship, Mike would be the winner with a [#permalink]

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18 Jul 2015, 10:03

FireStorm wrote:

katzzzz wrote:

Hi,

I mean..why can't we choose 6/7 as the answer. Probability of either one of them win = probability of mike lost the game? I not sure where did get it wrong.. Thanks

The most simple and direct answer to that question is the events are independent. I hope you get it now?

i thought that probability of mike winning =(Prob. of Ben not winning)*(Prob. of Rob not winning)*(Prob. of Mike winning)

FRom all explanations above it seems that championship does not have only 3 players But how can i identify this ?

Re: If Ben were to lose the championship, Mike would be the winner with a [#permalink]

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18 Jul 2015, 18:12

divya517 wrote:

FireStorm wrote:

katzzzz wrote:

Hi,

I mean..why can't we choose 6/7 as the answer. Probability of either one of them win = probability of mike lost the game? I not sure where did get it wrong.. Thanks

The most simple and direct answer to that question is the events are independent. I hope you get it now?

i thought that probability of mike winning =(Prob. of Ben not winning)*(Prob. of Rob not winning)*(Prob. of Mike winning)

FRom all explanations above it seems that championship does not have only 3 players But how can i identify this ?

Thanks

Hi,

You are looking at the wrong set of events.

There may or may not be other players, but we are asked to find out the probability of either Mike or Rob winning with respect to the probability of Ben losing, which is why we don't go about solving this the way you did [probability of mike winning =(Prob. of Ben not winning)*(Prob. of Rob not winning)*(Prob. of Mike winning)].

Hope it's clear.
_________________

I used to think the brain was the most important organ. Then I thought, look what’s telling me that.

The most simple and direct answer to that question is the events are independent. I hope you get it now?

i thought that probability of mike winning =(Prob. of Ben not winning)*(Prob. of Rob not winning)*(Prob. of Mike winning)

FRom all explanations above it seems that championship does not have only 3 players But how can i identify this ?

Thanks

Hi,

You are looking at the wrong set of events.

There may or may not be other players, but we are asked to find out the probability of either Mike or Rob winning with respect to the probability of Ben losing, which is why we don't go about solving this the way you did [probability of mike winning =(Prob. of Ben not winning)*(Prob. of Rob not winning)*(Prob. of Mike winning)].

Hope it's clear.

Hi, there have to be more players... otherwise divya517 & katzzzz are correct , the answer would be 6/7... But the Question is giving probability of the two player winning when ben does not win, and since the combined prob of either of them winning is not equal to ben losing, so there are more player who too can win if ben loses _________________

Re: If Ben were to lose the championship, Mike would be the winner with a [#permalink]

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18 Jul 2015, 22:25

Hi,

Yes, there are more. But strictly speaking that's a circuitous route of going around this problem. We are asked to find the probability only w.r.t Ben's chances of winning/losing.
_________________

I used to think the brain was the most important organ. Then I thought, look what’s telling me that.

If Ben were to lose the championship, Mike would be the winner with a probability of \(\frac{1}{4}\), and Rob - \(\frac{1}{3}\) . If the probability of Ben being the winner is \(\frac{1}{7}\), what is the probability that either Mike or Rob will win the championship? Assume that there can be only one winner.

A. \(\frac{1}{12}\) B. \(\frac{1}{7}\) C. \(\frac{1}{2}\) D. \(\frac{7}{12}\) E. \(\frac{6}{7}\)

M07-12

Case-1: Ben loses and Mile Wins--- The probability of Ben losing is 6/7 and then Mike winning is 1/4 so the total Probability of Case-1 = (6/7)*(1/4)

Case-2: Ben loses and Mile Wins--- The probability of Ben losing is 6/7 and then Rob winning is 1/3 so the total Probability of Case-1 = (6/7)*(1/3)

Total Probability of Case-1&2 together = (6/7)*(1/4)+(6/7)*(1/3) = 1/2

Answer: Option C
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