Question 4

Scenario: A certain system can experience certain types of defects.

Events: where for each type of defect.

Probabilities:

  • .
  • .
  • .
  • .
  • .
  • .
  • .

(a)

Problem: What is the probability that the system does not have a type 1 defect?

Process: The probability of the system not having a type 1 defect is the probability of the complement of .

Answer: 0.84.

(b)

Problem: What is the probability that the system has both type 1 and type 2 defects?

Process: The phrase "both type 1 and type 2 defects" indicates we're finding the probability of the intersection (and) of the events in and .

Answer: 0.08.

(c)

Problem: What is the probability that the system has both type 1 and type 2 defects but not type 3 defects?

Process: The phrase "both type 1 and type 2 defects but not type 3 defects" indicates we're finding the probability of the intersection (and) of the events in , and the complement of .

Answer: 0.06.

Summary

To find the probability that one event doesn't occur while two other events do occur:

(d)

Problem: What is the probability that the system has at most two of these defects?

Process: The phrase "has at most two of these defects" indicates we're finding the probability of the complement of the intersection (and) of , , and . The intersection of , , and represents the situation where the system has all three defects. So, the probability that the system has at most two types of defects is equivalent to the complement of the probability where it has all three types of defects. This probability is also equivalent to the probability of the union (or) of the complements of , , and . The union of the complements of , , and represents the situation where at least one of the defects doesn't occur.

Answer: 0.98.

Summary

To find the probability that at least one event doesn't occur if you know the probability that all three events occur: