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SVP
Joined: 16 Oct 2003
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John and Mary are taking a mathematics course. The course [#permalink]
 04 Nov 2003, 15:01
John and Mary are taking a mathematics course. The course has only three grades: A, B, and C. The probability that John gets a B is .3. The probability that Mary gets a B is .4. The probability that neither gets an A but at least one gets a B is .1. What is the probability that at least one gets a B but neither gets a C?
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SVP
Joined: 30 Oct 2003
Posts: 1963
Location: NewJersey USA
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P (Desired event ) = Neither gets C but atleast one gets B
Let us define the following
Ja = probability that John gets A
Jb = probability that John gets B
Jc = probability that John gets C
Ma = probability that Mary gets A
Mb = probability that Mary gets B
Mc = probability that Mary gets C
Desired event = Ma Jb + Mb Jb + Mb Ja
We also know that Ma+Mb+Mc = 1 and Ja+Jb+Jc = 1
So
Ma = 1- ( Mb + Mc ) and Ja = 1 - ( Jb + Jc )
P = ( 1- ( Mb + Mc ) ) * Jb + Mb Jb + ( 1 - ( Jb + Jc ) ) * Mb
P = Jb - Jb Mb - Jb Mc + Mb Jb + Mb - Jb Mb - Mb Jc
P = Jb + Mb - ( Jb Mc + Jb Mb + Jc Mb )
We also know that probability that neither gets A but atleast one gets B is 0.1 = Jb Mc + Jb Mb + Jc Mb
so P = 0.3+0.4 - 0.1 = 0.6
Answer is 0.6
I hope I am right. If some one can find a shorter method it will be great
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Senior Manager
Joined: 22 May 2003
Posts: 341
Location: Uruguay
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I also got 60%
Information given:
I) Jb=0.3 => Ja+Jc=0.7
II) Mb=0.4 => Ma+Mc=0.6
III) Jb*Mb + Jc*Mb + Jb*Mc = 0.1
We have to find:
IV) Jb*Mb + Ja*Mb + Jb*Ma = X
Calculations:
III+IV) 2(Jb*Mb) + Mb*(Ja+Jc) + Jb*(Ma+Mc) = 0.1 + X
2*0.3*0.4 + 0.4*0.7 + 0.3*0.6 = 0.1 + X
X=0.6
Martin
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SVP
Joined: 30 Oct 2003
Posts: 1963
Location: NewJersey USA
Followers: 3
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Hi Martin
I liked your approach better.
Anand.
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