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Mary and John are taking a mathematics course. This course has only.pdf

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Mary and John are taking a mathematics course. This 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? (Hint: let the sample space be the set of ordered pairs (John Solution 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.

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Mary and John are taking a mathematics course. This 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? (Hint: let the sample space be the set of ordered pairs (John Solution 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

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Mary and John are taking a mathematics course. This course has only.pdf

  • 1. Mary and John are taking a mathematics course. This 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? (Hint: let the sample space be the set of ordered pairs (John Solution 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