Probability Syntax: Types 1, 2 & 3 Defects

[SwaGGeReR]

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

P(A[1]) =.12
P(A[2])=.07
P(A[3])=.05
P(A[1] union A[2])=.13
P(A[1] union A[3])=.14
P(A[2] union A[3])=.10
P(A[1] intersects A[2] intersects A[3])=.01

1.what is the prob that the system does not have a type one defect?

2.what is the prob that the system has both type 1 and type 2 defects?

3. What is the prob that the system has both type 1 and type 2 defects but not a type 3 defect?

4.What is the prob that the system has at most two of these defects?

I know #1 is .88 and #4 is .99, but I am having difficulty understanding #2 and #3.

For #2: What is the probability that the system has both type 1 and type 2 defects, could that also include a system with ALL the defects (type 1, type 2, AND type 3 defects)? If so, I calculate the probability to be .07:

==>P(A int B) + P(A int B int C)
==>.12 + .07 - .13 + .01
==>.07

If so, then then the probability of #3 is .06.

Let me know if this is right or wrong.

 

[Stephen Tashi]

For #2: What is the probability that the system has both type 1 and type 2 defects, could that also include a system with ALL the defects (type 1, type 2, AND type 3 defects)?

Yes. The event A_1 \cap A_2 has the event A_1 \cap A_2 \cap A_3 as a subset.

If so, I calculate the probability to be .07:

==>P(A int B) + P(A int B int C)
==>.12 + .07 - .13 + .01
==>.07

By what you said above, you don't have to add the probability of (A int B in C) to the probability of A int B. The probability of A int B already accounts for the probability of A int B int C.

Also, I don't know how you calculated P(A int B).

Use the equation P(A_1 \cup A_2) = P(A_1) + P(A_2) - P(A_1 \cap A_2) and solve it for P(A_1 \cap A_2).