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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oeerivona oeerivona · Answer 1 · 2020-01-26T14:31:13+01:00

Answer:

Probability of either getting a B but neither gets a C = 0.6

Step-by-step explanation:

Probability of sought event = At least one gets a B but neither gets a C

Let

P = Probability of either getting a B but neither gets a C

P(JA)= The probability of getting an A by John

P(JB) = The probability of getting a B by John

P(JC) = The probability of getting a C by John

P(MA) = The probability of getting an A by Mary

P(MB) = The probability of getting a B by Mary

P(MC) = The probability of getting a C by Mary

Desired event = P(MA)×P(JB) + P(MB)× P(JB) + P(MB)× P(JA)

The probability of Mary having a grade is

P(MA) + P(MB) + P(MC) = 1 and P(JA) + P(JB) + P(JC) = 1

Rearranging

P(MA) = 1- ( P(MB) + P(MC) ) and P(JA) = 1 - ( P(JB) + P(JC) )

P = ( 1- ( P(MB) + P(MC) ) ) × P(JB) + P(MB)× P(JB) + ( 1 - ( P(JB) + P(JC) ) ) × P(MB)

P = P(JB) - P(JB)×P(MB) - P(JB)×P(MC) + P(MB)×P(JB) + P(MB) - P(JB)×P(MB) - P(MB)×P(JC)

P= P(JB)+ P(MB)-( P(JB)× P(MC)+ P(MB)×P(JB)+ P(MB)×P(JC))

We are told that the probability that neither gets an A but at least one gets B is

0.1 = P(JB)×P(MC) + P(JB)× P(MB) + P(JC)×P(MB)

Therefore the probability that at least one gets a B but neither gets a C

P = 0.3+0.4 - 0.1 = 0.6

Ans 0.6