Last year, Mike, Ben and Rob participated in a chess tournament that crowned 1 champion. If their respective probabilities of being crowned champion were 1/4, 1/3 and 1/6, then what's the probability that either Mike or Ben won?

A. 1/6
B. 5/12
C. 7/12
D. 5/6
E. 1


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I did it this way.
Here three contestant M, B and R has winning crown probability 1/4 , 1/3 and 1/6 respectively.
Only one participant will win the crown.
Now we have to calculate the probability of winning the crown either by M or by B. = MB'+M'B
If M win then B will be definitely loose the crown and vice versa. In such case B' = 2/3 and M'= 3/4
So. MB'+M'B= 1/4*2/3+3/4*1/3= 5/12.
B is answer.
Please tell me if I am wrong? And why?

the probability that either Mike or Ben won ; 1/4+1/3 ; 7/12
IMO C

[quote="Bunuel"]Last year, Mike, Ben and Rob participated in a chess tournament that crowned 1 champion. If their respective probabilities of being crowned champion were 1/4, 1/3 and 1/6, then what's the probability that either Mike or Ben won?

A. 1/6
B. 5/12
C. 7/12
D. 5/6
E. 1

[

The probability that either Mike or Ben won is 1/4 + 1/3 = 3/12 + 4/12 = 7/12.

Answer: C
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@banuel sir can you offer your 2 cents. Simply adding the probability of Mike And Ben winning means only these two are playing. How can we ignore Rob ? Having raised this questions to multiple quant experts all had the opinion that we cant simply add 1/4 and 1/3 . Instead P( M win x B lose x R lose ) + P (M lose x B win x R lose ) is what they recommended. Need your help.

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'
In this question, we are using the formula P(Mike or Ben) = P(Mike) + P(Ben) - P(both). Since it is stated that there can be only one champion, P(both) = 0 and thus, P(Mike or Ben) is simply the sum of the individual probabilities of success.
The reason we are ignoring Rob is that participation of Rob and his probability of success has no effect whatsoever in P(Mike or Ben). First, notice that Mike, Ben, and Rob are not the only participants in this tournament since 1/4 + 1/3 + 1/6 < 1. So not only we are ignoring Rob, we are ignoring any other participants as well. Next, the event that "Mike wins" implies that both Ben, Rob and any other participant loses. Similarly for the event "Ben wins". There are no events such as "Mike wins and Rob does not loose" or "Ben wins and Mike does not loose". That's why it is not necessary to multiply P(Mike wins) with P(Ben loses) or P(Rob loses).

In the 1-[Prob of Mike Win x Ben Lose x R lose + Prob of Mike Lose x Ben Win x R lose ] approach, you are not only assuming that Mike, Ben, and Rob are the only participants (which is not true) but also calculating as if it was possible that Mike wins and some other participant does not loose.
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