Question

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
P(A1) = .12 P(A2) = .07 P(A3) = .05

P(A1 U A2) = .13 P(A1 U A3) = .14

P(A2 U A3) = .10 P(A1 ∩ A2 ∩ A3) = .01

a) What is the probability that the system does not have a type 1 defect? (If this is A') then I don't need help on part a

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?

Answer 1

Answer:

a) 0.88

b) 0.06

c) 0.05

d) 0.99

Step-by-step explanation:

a) Yes you are right, Probability that the system does not have a type 1 defect = P(A1').

It can be calculated using the rule of probability:

   P(A1') = 1 - P(A1)

             = 1 - 0.12

   P(A1') = 0.88

b) The probability that the system has both type 1 and type 2 defects can be calculated by the following formula:

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

By rearranging the above-mentioned equation, we get:

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

                   = 0.12 + 0.07 - 0.13

P(A1 ∩ A2) = 0.06

c) Here we need to find the probability that the system has both type 1 and type 2 defects (A1 ∩ A2) but not a type 3 defect which means we need the probability P (A1 ∩ A2 ∩ A3')

For that, we can use the law of probability:

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

By rearranging the above-mentioned equation, we get:

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

                            = 0.06 - 0.01

P(A1 ∩ A2 ∩ A3') = 0.05

d) To find out the probability that the system has at most two of these defects, we will use the probability that all 3 defects are present and subtract it from the total probability (i.e. 1).

P(the system has at most two defects) = 1 - P(the system has all three defects)

We know that the probability of the system having all three defects P(A1 ∩ A2 ∩ A3) = 0.01

So, P(the system has all three defects) = 1 - 0.01

     P(the system has all three defects) = 0.99