A certain system can experience three different types of defects. Let denote the event that the system has a defect of type i. Suppose that
P(A1) = .12 P(A2) = .07 P(A3) = .05
P(A1∪A2) = .13 P(A1∪A3) = .14
P(A2∪A3) = .10 P(A1∩A2∩A3) = .01
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?
To solve these problems, we can use set theory and basic probability principles. Let's go through each question step by step:
a. What is the probability that the system does not have a type 1 defect?
To find the probability of not having a type 1 defect, we can calculate the complement of having a type 1 defect (denoted as A1').
P(A1') = 1 - P(A1) = 1 - 0.12 = 0.88
So, the probability that the system does not have a type 1 defect is 0.88.
b. What is the probability that the system has both type 1 and type 2 defects?
To find the probability of having both type 1 and type 2 defects, we can calculate the intersection (denoted as A1∩A2) of events A1 and A2.
P(A1∩A2) = 0.01
So, the probability that the system has both type 1 and type 2 defects is 0.01.
c. What is the probability that the system has both type 1 and type 2 defects but not a type 3 defect?
To find this probability, we need to calculate the intersection of A1 and A2 (both defects present) and take the complement of A3 (type 3 defect not present). Let's denote this event as (A1∩A2')∩A3'.
P((A1∩A2')∩A3') = P(A1) - P(A1∩A2∩A3) = 0.12 - 0.01 = 0.11
So, the probability that the system has both type 1 and type 2 defects but not a type 3 defect is 0.11.
d. What is the probability that the system has at most two of these defects?
This question requires finding the union of all possible combinations of events where at most two defects occur (A1∪A2, A1∪A3, and A2∪A3). Let's denote this event as (A1∪A2∪A3).
P(at most two defects) = P(A1∪A2∪A3)
To find this probability, we can use the inclusion-exclusion principle:
P(A1∪A2∪A3) = P(A1) + P(A2) + P(A3) - P(A1∩A2) - P(A1∩A3) - P(A2∩A3) + P(A1∩A2∩A3)
Substituting the given values, we have:
P(at most two defects) = 0.12 + 0.07 + 0.05 - 0.13 - 0.14 - 0.10 + 0.01
P(at most two defects) = 0.08
So, the probability that the system has at most two of these defects is 0.08.