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If Ben does not win the championship, the probability of Mike winning is \(\frac{1}{4}\), while the probability of Rob winning is \(\frac{1}{3}\). Given that the probability of Ben winning is \(\frac{1}{7}\), what is the probability of either Mike or Rob winning the championship, assuming that only one individual can win?
A. \(\frac{1}{12}\)
B. \(\frac{1}{7}\)
C. \(\frac{1}{2}\)
D. \(\frac{7}{12}\)
E. \(\frac{6}{7}\)
M07-12
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A. \(\frac{1}{12}\)
B. \(\frac{1}{7}\)
C. \(\frac{1}{2}\)
D. \(\frac{7}{12}\)
E. \(\frac{6}{7}\)
M07-12
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Official Solution:
If Ben does not win the championship, the probability of Mike winning is \(\frac{1}{4}\), while the probability of Rob winning is \(\frac{1}{3}\). Given that the probability of Ben winning is \(\frac{1}{7}\), what is the probability of either Mike or Rob winning the championship, assuming that only one individual can win?
A. \(\frac{1}{12}\)
B. \(\frac{1}{7}\)
C. \(\frac{1}{2}\)
D. \(\frac{7}{12}\)
E. \(\frac{6}{7}\)
This problem involves the concept of conditional probability. It is important to understand that the direct probabilities of Mike and Rob winning the championship are not provided. Instead, we are given the probabilities of Mike and Rob winning the championship, only under the condition that Ben does not win. This means that the probabilities of Mike and Rob winning (\(\frac{1}{4}\) and \(\frac{1}{3}\), respectively) are relevant only if we consider the scenario where Ben does not win the championship.
The probability of Ben losing is \(1 - \frac{1}{7} = \frac{6}{7}\). 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 the probability of either Mike or Rob winning the championship is \(P = \frac{6}{7}(\frac{1}{4} + \frac{1}{3}) = \frac{1}{2}\).
Alternatively, consider the following:
Take 84 championship cases (84 is chosen as it is the least common multiple of 3, 4, and 7).
Out of these 84 cases, Ben will lose in \(\frac{6}{7} *84 = 72\). When Ben loses, Mike would win in \(72*\frac{1}{4} = 18\) cases (which is one-fourth of the cases when Ben loses), and Rob would win in \(72*\frac{1}{3} = 24\) cases. Therefore, the probability of either Mike or Rob winning the championship is given by \(P = \frac{18 + 24}{84} = \frac{1}{2}\).
Answer: C
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If Ben does not win the championship, the probability of Mike winning is \(\frac{1}{4}\), while the probability of Rob winning is \(\frac{1}{3}\). Given that the probability of Ben winning is \(\frac{1}{7}\), what is the probability of either Mike or Rob winning the championship, assuming that only one individual can win?
A. \(\frac{1}{12}\)
B. \(\frac{1}{7}\)
C. \(\frac{1}{2}\)
D. \(\frac{7}{12}\)
E. \(\frac{6}{7}\)
This problem involves the concept of conditional probability. It is important to understand that the direct probabilities of Mike and Rob winning the championship are not provided. Instead, we are given the probabilities of Mike and Rob winning the championship, only under the condition that Ben does not win. This means that the probabilities of Mike and Rob winning (\(\frac{1}{4}\) and \(\frac{1}{3}\), respectively) are relevant only if we consider the scenario where Ben does not win the championship.
The probability of Ben losing is \(1 - \frac{1}{7} = \frac{6}{7}\). 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 the probability of either Mike or Rob winning the championship is \(P = \frac{6}{7}(\frac{1}{4} + \frac{1}{3}) = \frac{1}{2}\).
Alternatively, consider the following:
Take 84 championship cases (84 is chosen as it is the least common multiple of 3, 4, and 7).
Out of these 84 cases, Ben will lose in \(\frac{6}{7} *84 = 72\). When Ben loses, Mike would win in \(72*\frac{1}{4} = 18\) cases (which is one-fourth of the cases when Ben loses), and Rob would win in \(72*\frac{1}{3} = 24\) cases. Therefore, the probability of either Mike or Rob winning the championship is given by \(P = \frac{18 + 24}{84} = \frac{1}{2}\).
Answer: C
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General Discussion
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
(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
Barkatis
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)
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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.
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.
Barkatis
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}
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asmit123
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
C
asmit123
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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Bunuel
Is it Explicitly mentioned that there is a possibility of more than 3 players ? If not - Possibility of Mike or Rob winning would be prob of Ben not winning - 1\(-1/6 = 6/7\)
Lets look at LCM Approach - We know Ben Lost 72 Cases of which Mike and Rob managed to win 42 Cases - if there is no 3rd person then who is winning these 72 - 42 Cases.
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hD13
The wording of the question is a bit unusual, so I wouldn't worry too much about how to interpret it (an official question testing the same concepts would be worded a bit differently), but the only way the two fractions "1/3" and "1/4" in the question could be correct is if there was at least one other competitor not mentioned in the question. If 6/7 were the correct answer here, then 1/3 + 1/4 would need to equal 6/7, which is not true.
There's no need for the question to explicitly mention that there are other competitors. A question asking the following would be perfectly fair: "Ali has a 1/3 probability of winning the Gold Medal at pole vault, and Bimal has a 1/4 probability of winning the Gold Medal at pole vault. What is the probability either Ali or Bimal win the Gold Medal?" There must be one or more other competitors in this question, because 1/3 + 1/4 does not equal 1.
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If either Mike or Rob wins the championship, Ben loses it. No matter who among Mike and Rob wins, Ben must lose.
The probability that Ben loses is 6/7. How come this is not the same as Mike or Rob winning, given one person wins the championship?
The probability that Ben loses is 6/7. How come this is not the same as Mike or Rob winning, given one person wins the championship?
Barkatis
VP1030
Because there are other competitors in the race. The probability of one of the three (Ben, Mike, or Rob) winning is:
1/7 + 6/7 * (1/4 + 1/3) = 9/14.
Since this total is less than 1, it indicates that there must be other participants in the race.
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