asmit123 wrote:
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 \(P=\frac{18+24}{84}=\frac{1}{2}\).
Answer: C.
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