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

Dear Bunuel,
Here is my solution,
P(Mike win) = 6/7*1/4*2/3
P(Rob win) = 6/7*1/3*3/4
=> P(Mike win or Rob win) = 6/7*1/4*2/3 + 6/7*1/3*3/4 = 5/14
How is my solution wrong?
I don't understand why P=6/7(1/4+1/3), The question is either Mike or Rob will win
Please help me.

Official Solution:

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


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

If Ben were to lose the championship, Mike would be the winner with a probability of \(\frac{1}{4}\), and Rob - \(\frac{1}{3}\)
I interpreted the red marked area as
Probability(B(not win))*P(M(win))=1/4
Probability(B(not win))*P(R(win))=1/3

Any one pls help me understand the above sentence the significance of "If Ben were to lose the championship"

Bunuel
Are we not considering the probability(1-1/3) of Rob losing the game when we are taking the winning probability of Mike since Mike's winning probability already include all the possibilities, even that of Rob losing the game?
Is my thinking correct?

Bunuel and chetan2u

I did this problem with a scenario LWL+LLW and got an answer which was not in the list.
Could you please suggest where did I go wrong?

Regards,
Arup

Hi..

The winning of the other two, M and R, is related to losing of B and nothing else...
When you are taking LWL or LLW, you are making each case, say of M, related to R and B, and that of R, related to M and B..

The question just means that instead of total events as 1, the total events has become 6/7, so scenario is (L of B)(W of M)+(L of B)(W of R) = \(\frac{6}{7}*\frac{1}{4}+\frac{6}{7}*\frac{1}{3}=\frac{3}{14}+\frac{2}{7}=\frac{7}{14}=\frac{1}{2}\)

If Ben were to lose the championship, Mike would be the winner with a probability of 1/4,

what is the probability that either Mike or Rob will win the championship?

Why should it not be 1/4 +1/3 ?
Why i need to consider Ben lose again with probability which is already has winning chances of single person.