KillerSquirrel wrote:

cocosdreamland wrote:

i got 7/12

1/4 + 1/3 + 1/6 = 9/12

Mike or Bob win the championship

1/4 + 1/3 = 3/12 + 4/12 = 7/12

(C)

Well done cocosdreamland ! but you forgot to subtract the event for Ben winning.

p(A or B but not C) = p(A) + p(B) - p(C)

Mike = 1/4 = 3/12

Rob = 1/3 = 4/12

Ben = 1/6 = 2/12

3/12+4/12 - 2/12 = 5/12

the answer is (B)

Interesting!

but my question is...

since the probability of each's winning is

Mike 3/12

Rob 4/12

Ben 2/12

OTHERS 3/12

so isn't the probability of either Mike or Rob winning but not Ben and Others = Prob. of Mike + Prob. of Rob, which is 3/12 + 4/12?

and accroding to your explaination, if P (A) and P(B) already does not include P(C), then why is P(C) being substracted from P(A) + P(B)?

am I missing something here?