Question

Prob-4 Pressurized air are being used in a manufacturing company in its daily operation. There are three compressors in service; two-screw type compressors, (S1 ,S2) and one-reciprocal type compressor,(R). Each compressor, at any given time, is either off (0), or on (1) . Assume that 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. Assume that all outcomes in event B are equally likely. If P(A∩B)=45% and P(A∪B)=93%. Compute the probability that all compressors are off

Answer 1

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!!!