The probability that a washing machine will operate satisfactorily for 3 years is 0.82 while the probability that a floor polisher will operate satisfactorily over the same period of time is 0.78. Find the probability that in a 3-year period, only the floor polisher will operate satisfactorily.
I already got, P(A and B) = 0.78(0.82) = 0.6396
I'm not sure if I should subtract it with 0.78 or 0.82
subtraction does not enter the picture
So you want:
Prob(floor polisher ok AND the washing machine not ok)
prob(washing machine not ok) = 1 - .82 = .18
so Prob(floor polisher ok AND the washing machine not ok)
= (.78)(.18)
= .1404
Oh okay.
I see. Thanks a lot!
To find the probability that only the floor polisher will operate satisfactorily over a 3-year period, you need to subtract the probability that both the washing machine and the floor polisher will operate satisfactorily from the probability that only the floor polisher will operate satisfactorily.
Let's define events:
A: The washing machine operates satisfactorily over a 3-year period.
B: The floor polisher operates satisfactorily over a 3-year period.
You have already calculated P(A and B) to be 0.78 * 0.82 = 0.6396.
To find the probability that only the floor polisher will operate satisfactorily, we can subtract P(A and B) from the probability that only B occurs.
To find the probability that only B occurs, we need to find the complement of A, which is the probability that the washing machine does not operate satisfactorily. This can be calculated as 1 - P(A).
Given the probability that A occurs is 0.82, we can calculate P(not A) = 1 - P(A) = 1 - 0.82 = 0.18.
Therefore, the probability that only the floor polisher will operate satisfactorily is P(B) - P(A and B) = 0.78 - 0.6396 = 0.1404.
So, the probability that in a 3-year period only the floor polisher will operate satisfactorily is 0.1404 or 14.04%.