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02 Oct 2010, 06:57
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
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Question Stats:
55% (02:01) correct
45%
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wrong
based on 477
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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
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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26 Jan 2015, 02:57
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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
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
General Discussion
02 Oct 2010, 07:00
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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
