A certain system can experience three different types of defects. Let Ai (i = 1,2,3) denote the event that the system has a defect of type i. Suppose that the following probabilities are true.

P (A1) = 0.14

P (A2) = 0.10

P (A3) = 0.07

P (A1 ∪ A2) = 0.16

P (A1 ∪ A3) = 0.17

P (A2 ∪ A3) = 0.14

P (A1 ∩ A2 ∩ A3) = 0.02

a. What is the probability that the system does not have a type 1 defect? b. What is the probability that the system has both type 1 and type 2 defects? c. What is the probability that the system has both type 1 and type 2 defects but not a type 3 defect? d. What is the probability that the system has at most two of these defects?

a. 0.86

b. 0.08

c. 0.06

d. 0.98

Step-by-step explanation:

a.

The probability that system doesn't have type 1 defect is

P (A1') = 1-P (A1) = 1-0.14=0.86

Thus, the probability that system doesn't have type 1 defect is 86%.

b.

The probability that system have both type 1 defect and type 2 defect

P (A1∩A2) = P (A1) + P (A1) - P (A1∪A2)

P (A1∩A2) = 0.14+0.1-0.16=0.08

Thus, the probability that system have both type 1 defect and type 2 defect is 8%.

c.

The probability that system have both type 1 defect and type 2 defect but not type 3 defect

P (A1∩A2∩A3') = P (A1∩A2) - P (A1∩A2∩A3)

P (A1∩A2∩A3') = 0.08-0.02=0.06

Thus, the probability that system have both type 1 defect and type 2 defect but not type 3 defect is 6%.

d.

The probability that system has at most 2 defectives

P (At most 2 defectives) = 1-P (3 defectives)

P (3 defectives) = P (A1∩A2∩A3) = 0.02

P (At most 2 defectives) = 1-P (3 defectives) = 1-0.02=0.98

The probability that system has at most 2 defectives is 98%.