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If Ben were to lose the championship (m07q12) [#permalink]

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12 Feb 2009, 23:14

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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?

Re: If Ben were to lose the championship (m07q12) [#permalink]

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18 Sep 2009, 15:35

While I understand the explanation, why is 1- 1/7 or 6/7 incorrect?

If there are only three people in the race and the probability of one person winning is 1/7, the probability of "either" of the remaining two winning is 1-1/7. Why is this thinking wrong?

Re: If Ben were to lose the championship (m07q12) [#permalink]

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30 Sep 2009, 13:34

sid3699 wrote:

hgp2k wrote:

sid3699 wrote:

While I understand the explanation, why is 1- 1/7 or 6/7 incorrect?

If there are only three people in the race and the probability of one person winning is 1/7, the probability of "either" of the remaining two winning is 1-1/7. Why is this thinking wrong?

I have the same question.

Can someone please explain this?

There is more than 3 people in the race. The explanation provided at first is correct.

This question needs editing. There is a dash/minus in front of the probability of Rob that should be erased. Yes, probability cant be negative, so its obvious that its a typo. But it is still a typo. It made me lose one minute.
_________________

Re: If Ben were to lose the championship (m07q12) [#permalink]

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15 Apr 2010, 10:58

I have another question. As I understand the author has two terms: 1. If Ben were to lose the championship, Mike would be the winner with a probability of 1/4, and Rob - 1/3

It's mean that probability of Ben's winning = 0.

2. If the probability of Ben being the winner is 1/7, what is the probability that either Mike or Rob will win the championship?

And here Ben's probability is 1/7 and it's means that probability of Mike's or Ron's winning should also change. And we cannot use here the probabilities from first term.

1/4 = 6/7 * m* (1-r) where m and r are prob of mike and rob winning. 1/3 = 6/7 * (1-m) * r

should be

1/4 = 1* m* (1-r) where m and r are prob of mike and rob winning. 1/3 = 1* (1-m) * r

Bunuel m I correct?

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?

A. \(\frac{1}{12}\)

B. \(\frac{1}{7}\)

C. \(\frac{1}{2}\)

D. \(\frac{7}{12}\)

E. \(\frac{6}{7}\)

This is conditional probability question. We need the probability that either Mike or Rob will win the championship. So Ben must loose --> the probability of Ben loosing 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 \(P=\frac{6}{7}(\frac{1}{4}+\frac{1}{3})=\frac{1}{2}\).

Answer: C.

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 Ben will loose 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 loose) and Rob would be the winner \(72*\frac{1}{3}=24\) --> \(P=\frac{18+24}{84}=\frac{1}{2}\).

Re: If Ben were to lose the championship (m07q12) [#permalink]

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17 Nov 2010, 07:34

Bunuel wrote:

gurpreetsingh wrote:

I got it..

1/4 = 6/7 * m* (1-r) where m and r are prob of mike and rob winning. 1/3 = 6/7 * (1-m) * r

should be

1/4 = 1* m* (1-r) where m and r are prob of mike and rob winning. 1/3 = 1* (1-m) * r

Bunuel m I correct?

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?

A. \(\frac{1}{12}\)

B. \(\frac{1}{7}\)

C. \(\frac{1}{2}\)

D. \(\frac{7}{12}\)

E. \(\frac{6}{7}\)

This is conditional probability question. We need the probability that either Mike or Rob will win the championship. So Ben must loose --> the probability of Ben loosing 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 \(P=\frac{6}{7}(\frac{1}{4}+\frac{1}{3})=\frac{1}{2}\).

Answer: C.

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 Ben will loose 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 loose) and Rob would be the winner \(72*\frac{1}{3}=24\) --> \(P=\frac{18+24}{84}=\frac{1}{2}\).

Re: If Ben were to lose the championship (m07q12) [#permalink]

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17 Nov 2010, 11:01

sid3699 wrote:

While I understand the explanation, why is 1- 1/7 or 6/7 incorrect?

If there are only three people in the race and the probability of one person winning is 1/7, the probability of "either" of the remaining two winning is 1-1/7. Why is this thinking wrong?

I had the same question as you did, but Bunuel gave a good explanation.
_________________

Re: If Ben were to lose the championship (m07q12) [#permalink]

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17 Nov 2010, 12:01

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My first post-- it seems like this it a great questions to estimate (how I got it right anyway).

How I thought it through:

If when Ben loses, the probability either Mike or Rob will win is 7/12, then if Ben's probability to win is only 1/7, then Ben losing isn't helping Mike or Rob that much. That means that the overall probability that either Mike or Rob will win is only a little less than 7/12. C (1/2) is the only answer that fits, since A and B are too small, E is too large, and D doesn't make sense given that we know the probability is less than 7/12.

Thanks for the mathematical explanation guys, but it's always good to have an estimation strategy in your back pocket too! (Especially if you're bad at probability like me)

Re: If Ben were to lose the championship (m07q12) [#permalink]

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18 Nov 2010, 05:40

Welcome to GMAT Club!

We hope you achieve a great score you're aiming at!

mikeepeck wrote:

My first post-- it seems like this it a great questions to estimate (how I got it right anyway).

How I thought it through:

If when Ben loses, the probability either Mike or Rob will win is 7/12, then if Ben's probability to win is only 1/7, then Ben losing isn't helping Mike or Rob that much. That means that the overall probability that either Mike or Rob will win is only a little less than 7/12. C (1/2) is the only answer that fits, since A and B are too small, E is too large, and D doesn't make sense given that we know the probability is less than 7/12.

Thanks for the mathematical explanation guys, but it's always good to have an estimation strategy in your back pocket too! (Especially if you're bad at probability like me)

Now given,P(mike wins)=1/4.Therefore,P(mike looses)=3/4 Similarly,P(rob wins)=1/3.Therefore,P(rob looses)=2/3 And,P(ben wins)=1/7.Therefore,P(ben looses)=6/7

Putting the values, 6/7.1/4.2/3 + 6/7.3/4.1/3 = 1/7+3/14=5/14

please explain where I am going wrong.

If there exists a dependancy between the respective probabilities of two event, we multiply the two probabilities: Probability of Mike winning depends on Probability of Ben loosing, therefore P(M) = (6/7)*(1/4) Probability of Rob winning depends on Probability of Ben loosing, therefore P(R) = (6/7)*(1/3)

These two probability are mutually independent, therefore (this is where you went wrong): P(total) = P(M) + P(R) = (6/7)*[1/4 + 1/3] = 6/7 * 7/12 = 1/2
_________________

Re: If Ben were to lose the championship (m07q12) [#permalink]

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05 Jul 2011, 06:00

AriBenCanaan wrote:

sap wrote:

see this. Probability of of either winning = either 1 + either 2 - both 1 and 2 when we take OR case, we need to subtract where both of them win. so the solution becomes 1/4+1/3-1/12 = 1/2

I think the above solution is incorrect since the probability of both 1 & 2 = 0 in this case since there is only one winner. Can someone help confirm this?

Also, why is the probability of Mike wining (1/4) or the probability of Rob wining (1/3) not an absolute probability and rather a probability of them wining provided Ben loses (From Bunuel's solution)? So basically, why do we need to multiply the probability of Ben losing when its implied that when Mike or Rob win, Ben has to lose?

This may seem like a silly question to the math gurus, but I've spent a lot of time on this and can't seem to figure this piece out.....

A couple of points to remember -- the question is asking for the probability of "EITHER" Mike "OR" Rob winning. It is not asking for the probability that both win ("AND"). They are mutually exclusive events (i.e Mikes wins = Rob does not win). Two events are mutually exclusive if they cannot occur at the same time.

When you see an "OR" question -- you always add. When you see an "AND" question you multiply (Independent events).

So it comes down to two categories -- 1) Independent events 2) Mutually exclusive events

If you understand the difference between Independent events and Mutually exclusive events -- then the solution to this problem will make a lot more sense. MGMAT has a nice section of this stuff. There is some good stuff online as well -- that really simplifies these concepts.

Re: If Ben were to lose the championship (m07q12) [#permalink]

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09 Oct 2011, 17:02

Bunuel wrote:

gurpreetsingh wrote:

I got it..

1/4 = 6/7 * m* (1-r) where m and r are prob of mike and rob winning. 1/3 = 6/7 * (1-m) * r

should be

1/4 = 1* m* (1-r) where m and r are prob of mike and rob winning. 1/3 = 1* (1-m) * r

Bunuel m I correct?

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?

A. \(\frac{1}{12}\)

B. \(\frac{1}{7}\)

C. \(\frac{1}{2}\)

D. \(\frac{7}{12}\)

E. \(\frac{6}{7}\)

This is conditional probability question. We need the probability that either Mike or Rob will win the championship. So Ben must loose --> the probability of Ben loosing 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 \(P=\frac{6}{7}(\frac{1}{4}+\frac{1}{3})=\frac{1}{2}\).

Answer: C.

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 Ben will loose 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 loose) and Rob would be the winner \(72*\frac{1}{3}=24\) --> \(P=\frac{18+24}{84}=\frac{1}{2}\).

Answer: C.

Hope it's clear.

Bunuel - I have a question -

In your explanation, Out of 84 people, Ben will lose in 12 cases. Therefore, we have 72 possibilities open for either Mike or Rob.

Now out of 72 cases, Mike would win in 18 and 24 cases respectively. My question is that what would happen to the remaining 84-12-18-24 = 30 cases? I am not sure whether I understood the break-up correctly.

Re: If Ben were to lose the championship (m07q12) [#permalink]

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09 Feb 2012, 12:38

sid3699 wrote:

While I understand the explanation, why is 1- 1/7 or 6/7 incorrect?

If there are only three people in the race and the probability of one person winning is 1/7, the probability of "either" of the remaining two winning is 1-1/7. Why is this thinking wrong?

Re: If Ben were to lose the championship (m07q12) [#permalink]

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05 Nov 2013, 07:50

Bunuel wrote:

gurpreetsingh wrote:

I got it..

1/4 = 6/7 * m* (1-r) where m and r are prob of mike and rob winning. 1/3 = 6/7 * (1-m) * r

should be

1/4 = 1* m* (1-r) where m and r are prob of mike and rob winning. 1/3 = 1* (1-m) * r

Bunuel m I correct?

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?

A. \(\frac{1}{12}\)

B. \(\frac{1}{7}\)

C. \(\frac{1}{2}\)

D. \(\frac{7}{12}\)

E. \(\frac{6}{7}\)

This is conditional probability question. We need the probability that either Mike or Rob will win the championship. So Ben must loose --> the probability of Ben loosing 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 \(P=\frac{6}{7}(\frac{1}{4}+\frac{1}{3})=\frac{1}{2}\).

Answer: C.

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 Ben will loose 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 loose) and Rob would be the winner \(72*\frac{1}{3}=24\) --> \(P=\frac{18+24}{84}=\frac{1}{2}\).

Answer: C.

Hope it's clear.

Out of 84 only 24 and 18 are won = 42, How come?
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