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# If the probability that Mike can win a championship is 1/4,

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Director
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If the probability that Mike can win a championship is 1/4, [#permalink]

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17 Jul 2007, 15:32
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If the probability that Mike can win a championship is 1/4, and that of Rob winning is 1/3 and that of Ben winning is 1/6, what is the probability that Mike or Ben will win the championship but not Ben?

a) 1/6
b) 5/12
c) 7/12
d) 5/6
e) 1

Manager
Joined: 07 Feb 2007
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17 Jul 2007, 19:08
GK_Gmat wrote:
If the probability that Mike can win a championship is 1/4, and that of Rob winning is 1/3 and that of Ben winning is 1/6, what is the probability that Mike or Ben will win the championship but not Ben?

a) 1/6
b) 5/12
c) 7/12
d) 5/6
e) 1

I am confused about "Mike or Ben will win the championship but not Ben? "
Intern
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17 Jul 2007, 19:24
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)
Director
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17 Jul 2007, 19:48
(M and R and ~B) + (M and ~R and ~B) + (~M and R and ~B)

= 1/4 * 1/3* 5/6 + 1/4 * 2/3 * 5/6 + 3/4 * 1/3 * 5/6

= 5/6 * 1/3 [ 1/4 + 2/4 + 3/4]
= 5/6 * 1/3 * 6/4
= 5/12
VP
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17 Jul 2007, 21:17
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

Intern
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17 Jul 2007, 21:38
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

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?
VP
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17 Jul 2007, 21:42
cocosdreamland wrote:

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?

I don't think you are missing anything - this is a poorly written question that was discussed in this forum many times.

Director
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19 Jul 2007, 23:42
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

Killer Squirrel can u pls. explain why we subtract the P(not C)?

Shouldn't this be read as p(A or B and not C) which would be

p(A) + p(B) * P(not C)??

Thanks
VP
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20 Jul 2007, 00:05
20 Jul 2007, 00:05
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