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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] New post 17 Jul 2007, 14:32
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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

Pls. explain your answer.
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Re: PS: Probability Champ [#permalink] New post 17 Jul 2007, 18:08
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

Pls. explain your answer.


I am confused about "Mike or Ben will win the championship but not Ben? "
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 [#permalink] New post 17 Jul 2007, 18:24
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)
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 [#permalink] New post 17 Jul 2007, 18:48
(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
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 [#permalink] New post 17 Jul 2007, 20:17
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

the answer is (B)

:-D
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 [#permalink] New post 17 Jul 2007, 20:38
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

the answer is (B)

:-D


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? :(
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 [#permalink] New post 17 Jul 2007, 20:42
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.

:-D
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 [#permalink] New post 19 Jul 2007, 22:42
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

the answer is (B)

:-D


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
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 [#permalink] New post 19 Jul 2007, 23:05
  [#permalink] 19 Jul 2007, 23:05
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