Answer:
The probability that all compressors are off = P(A ∩ B') = 0.18
Step-by-step explanation:
Event A is denoted to reciprocal compressor (R) that is always off.
Event B is denoted that at least one of the screw type of compressor is always on.
P(A) = Probability that the reciprocal compressor is off.
P(A') = Probability that the reciprocal compressor is on.
P(B) = Probability that at least one of the screw type of compressor is on.
P(B') = Probability that at least one of the screw type of compressor is off.
P(A ∩ B) = 45% = 0.45
P(A ∪ B) = 93% = 0.93
P(U) = P(A ∪ B) + P(A' ∩ B')
1 = 0.93 + P(A' ∩ B')
P(A' ∩ B') = 1 - 0.93 = 0.07
The probability that all compressors are off is given as P(A ∩ B')
P(A) = P(A n B') + P(A n B)
P(B) = P(A n B) + P(A' n B)
The question asks us to assume that all outcomes in event B are equally likely.
The possible outcomes in event B include
- The two compressors are on
- First compressor is on, second compressor is off
- Second compressor is on, first compressor is off
- both compressors are off
Since all the outcomes are equally likely, the probability that at least one of the two compressors is on = (3/4) = 0.75 = P(B)
P(B) = P(A n B) + P(A' n B)
0.75 = 0.45 + P(A' n B)
P(A' n B) = 0.75 - 0.45 = 0.30
P(A ∪ B) = P(A ∩ B') + P(A' ∩ B) + P(A ∩ B)
0.93 = P(A ∩ B') + 0.30 + 0.45
P(A ∩ B') = 0.93 - 0.30 - 0.45 = 0.18
Hope this Helps!!!
