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Re: If Ben were to lose the championship (m07q12) [#permalink]
05 Nov 2013, 06:53
Expert's post
honchos wrote:
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?
Sorry, I don't get your question at all... Please elaborate. Thank you. _________________
Re: If Ben were to lose the championship (m07q12) [#permalink]
15 Nov 2013, 23:43
Please answer the question asked by voodoochild
voodoochild wrote:
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]
15 Nov 2013, 23:45
1
This post received KUDOS
Probability tree is very useful to solve this kind of question. C is the answer. Here is my diagram. Hope it helps
Attachments
Untitled.png [ 15.43 KiB | Viewed 784 times ]
_________________
Please +1 KUDO if my post helps. Thank you.
"Designing cars consumes you; it has a hold on your spirit which is incredibly powerful. It's not something you can do part time, you have do it with all your heart and soul or you're going to get it wrong."
Re: If Ben were to lose the championship (m07q12) [#permalink]
15 Nov 2013, 23:57
pqhai wrote:
Probability tree is very useful to solve this kind of question. C is the answer. Here is my diagram. Hope it helps
This is one of the very amazing way of solving, But i am still trying to grab the logic behinds Bunel's solution- Bunuel's approach- Take 84 championships/cases (I chose 84 as it's a LCM of 3, 4, and 7).
My Doubt: Out of 84 only 24 and 18(1/4th and 1/3rd) are won = 42, How come?
Even if we add 1/7 X 84 = 12. it becomes 54, what about other matches, is this question correct? _________________
Re: If Ben were to lose the championship (m07q12) [#permalink]
16 Nov 2013, 00:23
honchos wrote:
pqhai wrote:
Probability tree is very useful to solve this kind of question. C is the answer. Here is my diagram. Hope it helps
This is one of the very amazing way of solving, But i am still trying to grab the logic behinds Bunel's solution- Bunuel's approach- Take 84 championships/cases (I chose 84 as it's a LCM of 3, 4, and 7).
My Doubt: Out of 84 only 24 and 18(1/4th and 1/3rd) are won = 42, How come?
Even if we add 1/7 X 84 = 12. it becomes 54, what about other matches, is this question correct?
Hello honchos.
The logic behind Bunuel's approach is that if you have two variables having own probabilities ==> The probability of happening BOTH two variables = (Probability of variable 1) x (Probability of variable 2)
I will answer your first question: why the probability = 1/2 Let apply the logic above to the question: Mike win: Mike wins ONLY IF Ben loses ==> The probability of Mike wins MUST be = (Probability of Ben loses) x(probability of Mike wins) = (84*6/7)*1/4 = 72*1/4 = 18 Rob win: The same pattern is true for Rob ==> Probability of Rob wins = (Probability of Ben loses) x(probability of Rob wins) = (84*6/7)*1/3 = 72*1/4 = 24 ==> Total = 42 ==> Probability of either Mike or Rob win = 42/84 = 1/2
About your second question. Why 84*1/7 + 42 is only 54. As you thought, it SHOULD be 84? No, 84 = [probability of Ben loseWITHOUT the impact of Mike and Rob] + Probability of Ben win = 84*6/7 + 84*1/7
Hope it's clear. _________________
Please +1 KUDO if my post helps. Thank you.
"Designing cars consumes you; it has a hold on your spirit which is incredibly powerful. It's not something you can do part time, you have do it with all your heart and soul or you're going to get it wrong."
Re: If Ben were to lose the championship (m07q12) [#permalink]
16 Nov 2013, 00:34
1
This post received KUDOS
honchos wrote:
I d not agree with you- "No, 84 = [probability of Ben lose WITHOUT the impact of Mike and Rob] + Probability of Ben win = 84*6/7 + 84*1/7 "
84 is the total #(Sample Space) of matches thats the reason we chose 6/7 and 1/7 to calculate.
I think you may misunderstand the question a bit. You assume there are only three people Ben, Mike and Rob. So, if Ben loses, either Mike or Rob is automatically a winner? If that's the case, why probability of Mike's win and that of Rob are 1/4 and 1/3 respectively?
Do not assume that Ben loses = Mike or Rob is automatically a winner. Because if that's the case ==> The probability of Mike or Rob win is automatically = 6/7 = probability of Ben loses. That's wrong.
Hope it makes sense. _________________
Please +1 KUDO if my post helps. Thank you.
"Designing cars consumes you; it has a hold on your spirit which is incredibly powerful. It's not something you can do part time, you have do it with all your heart and soul or you're going to get it wrong."
Re: If Ben were to lose the championship (m07q12) [#permalink]
22 Nov 2013, 13:36
Sunita123, it is dash 1/3 and not -1/3. The probability of happening or non happening of an event lies between 0 and 1 both inclusive. So, probability can NEVER be negative or greater than 1.
Re: If Ben were to lose the championship (m07q12) [#permalink]
09 May 2014, 03:21
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 It is no where mentioned that events are mutually exclusive or noe, whats wrong in this method:
Re: If Ben were to lose the championship (m07q12) [#permalink]
09 May 2014, 04:02
Expert's post
honchos wrote:
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 It is no where mentioned that events are mutually exclusive or noe, whats wrong in this method:
6/7 X 1/4 X 2/3 + 6/7 X 3/4 X1/3
Mike winning automatically means Rob loosing. Similarly Rob winning automatically means Mike loosing. So, the assumption is that only one person can win the championship.
Mike would be the winner with a probability of \(\frac{1}{4}\) IF Ben were to lose: 1/4*6/7. Rob would be the winner with a probability of \(\frac{1}{3}\) IF Ben were to lose: 1/3*6/7.
So, no need to multiply above by the probability of Rob/Mike losing too. _________________