Question

John and Mary are taking a discrete mathematics course. The course has only three grades: A, B, and C. The probability that John gets a B is 0.3. The probability that Mary gets a B is 0.4. The probability that neither gets an A but at least one gets a B is 0.1. What is the probability that at least one gets a B but neither gets a C?

Answer 1

Answer:

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