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# John and Mary are taking a mathematics course. The course

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Joined: 16 Oct 2003
Posts: 1815
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John and Mary are taking a mathematics course. The course [#permalink]  04 Nov 2003, 14:01
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?
SVP
Joined: 30 Oct 2003
Posts: 1797
Location: NewJersey USA
Followers: 5

Kudos [?]: 41 [0], given: 0

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

I hope I am right. If some one can find a shorter method it will be great
Senior Manager
Joined: 22 May 2003
Posts: 334
Location: Uruguay
Followers: 1

Kudos [?]: 34 [0], given: 0

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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Joined: 30 Oct 2003
Posts: 1797
Location: NewJersey USA
Followers: 5

Kudos [?]: 41 [0], given: 0

Hi Martin

Anand.
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