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) = 0.34, P(A2) = 0.33, P(A3) = 0.43,
P(A1 ∪ A2) = 0.6, P(A1 ∪ A3) = 0.64,
P(A2 ∪ A3) = 0.67, P(A1 ∩ A2 ∩ A3) = 0.03
(a) Find the probability that the system has exactly 2 of the 3 types of defects.
(b) Find the probability that the system has a type 1 defect given that it does not have a type 2 or type 3 defect.
0.25
To solve these probability problems, we can use the principle of inclusion-exclusion. The principle of inclusion-exclusion states that for three events A, B, and C, the probability of the union of these events is given by:
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C)
We can use this principle to solve both parts of the problem.
(a) Finding the probability that the system has exactly 2 of the 3 types of defects:
Let's denote the event that the system has exactly 2 types of defects as E. We can express E in terms of union and intersection of events A1, A2, and A3 as follows:
E = (A1 ∩ A2 ∩ ¬A3) ∪ (A1 ∩ ¬A2 ∩ A3) ∪ (¬A1 ∩ A2 ∩ A3)
Now, let's calculate the probability of E using the principle of inclusion-exclusion:
P(E) = P(A1 ∩ A2 ∩ ¬A3) + P(A1 ∩ ¬A2 ∩ A3) + P(¬A1 ∩ A2 ∩ A3)
To calculate the probability of each term, we need to know the probabilities of the individual events and their intersections. We already have the probabilities of the individual events (P(A1), P(A2), P(A3)), as well as the probabilities of their unions (P(A1 ∪ A2), P(A1 ∪ A3), P(A2 ∪ A3)), and the probability of their intersection (P(A1 ∩ A2 ∩ A3)). We can use these values to calculate P(E).
P(A1 ∩ A2 ∩ ¬A3) = P(A1 ∪ A2) - P(A1 ∪ A2 ∪ A3)
P(A1 ∩ ¬A2 ∩ A3) = P(A1 ∪ A3) - P(A1 ∪ A2 ∪ A3)
P(¬A1 ∩ A2 ∩ A3) = P(A2 ∪ A3) - P(A1 ∪ A2 ∪ A3)
Substituting these values into the formula for P(E):
P(E) = (P(A1 ∪ A2) - P(A1 ∪ A2 ∪ A3)) + (P(A1 ∪ A3) - P(A1 ∪ A2 ∪ A3)) + (P(A2 ∪ A3) - P(A1 ∪ A2 ∪ A3))
Using the given values, we can substitute the known probabilities to calculate P(E).
(b) Finding the probability that the system has a type 1 defect given that it does not have a type 2 or type 3 defect:
Let's denote the event that the system has a type 1 defect as F and the event that it does not have a type 2 or type 3 defect as G. We want to find P(F|G), which represents the conditional probability of F given G.
Using the definition of conditional probability:
P(F|G) = P(F ∩ G) / P(G)
The probability of F ∩ G represents the probability that the system has both a type 1 defect and does not have a type 2 or type 3 defect. In other words, it represents the probability of the intersection of F and G.
We can calculate P(F ∩ G) by subtracting the probability of the system having a type 1 defect and both type 2 and type 3 defects from the probability of the system having a type 1 defect:
P(F ∩ G) = P(F) - P(F ∩ A2 ∩ A3)
Again, using the given values, we can substitute the known probabilities to calculate P(F ∩ G). Then, we can use P(F ∩ G) and the given probabilities of G (¬A2 ∩ ¬A3) to calculate P(F|G).
By following these steps, we can find the probabilities requested in the problem.