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. probabilities are:
P(A1) = 0.12 P(A2) = 0.07 P(A3) = 0.05
P(A1 union A2) = 0.13 P(A1 union A3) = 0.14
P(A2 union A3) = 0.10 P(A1 intersect A2 intersect A3) = 0.01
(a) Given that the system has a type 1 defect, what is the probability that it has a type 2 defect?
I did (0.12+0.07-0.13)/0.12 and got 0.5
(b) Given that the system has a type 1 defect, what is the probability that it has all three types of defects?
Not sure. I think I'm supposed to do 0.01/0.12 but I'm not positive
(c) Given that the system has at least one type of defect, what is the probability that it has exactly one type of defect?
Not sure
(d) Given that the system has both of the first two types of defects, what is the probability that it does not have the third type of defect?
Not sure.
I just need some help with b, c, and d. I think A is right and I don't know about the rest...
(a) Your answer for (a) is correct.
(b) You're correct in your approach for (b). The probability that it has all three types of defects given that it has a type 1 defect is P(A1 intersect A2 intersect A3) / P(A1) = 0.01 / 0.12 = 1/12 ≈ 0.083
(c) To find the probability of having exactly one type of defect given that the system has at least one type of defect, we first need to compute the probabilities of having exactly 1, 2, or 3 types of defects.
Let A1_only be the event that the system has only a type 1 defect, A2_only be the event that it has only a type 2 defect, and A3_only be the event that it has only a type 3 defect.
We find the probabilities of those events as follows:
P(A1_only) = P(A1) - P(A1 union A2) - P(A1 union A3) + P(A1 intersect A2 intersect A3) = 0.12 - 0.13 - 0.14 + 0.01 = 0.00-0.14 = -0.02 (meaning that there are no cases when only A1 is present; in all cases, A1 comes with at least one of the other defects)
P(A2_only) = P(A2) - P(A1 union A2) - P(A2 union A3) + P(A1 intersect A2 intersect A3) = 0.07 - 0.13 - 0.10 + 0.01 = -0.15
P(A3_only) = P(A3) - P(A1 union A3) - P(A2 union A3) + P(A1 intersect A2 intersect A3) = 0.05 - 0.14 - 0.10 + 0.01 = -0.08
Sum the three probabilities: P(one_defect) = P(A1_only) + P(A2_only) + P(A3_only) = 0
The given probabilities seem to result in no cases when there is only one defect, which is counterintuitive. There might be some error in the problem's data.
(d) Given that the system has both of the first two types of defects, we need to find the probability that it does not have the third type of defect. In other words, we're looking for P(A3' | A1 intersect A2), where A3' is the complement of A3 (meaning the system does not have a type 3 defect).
By conditional probability, P(A3' | A1 intersect A2) = P(A3' intersect (A1 intersect A2)) / P(A1 intersect A2).
We know P(A1 intersect A2 intersect A3) = 0.01, so P(A1 intersect A2) = P(A1 union A2) - P(A3 | A1 union A2) = 0.13 - 0.01 = 0.12.
Now we need to find P(A3' intersect (A1 intersect A2)). Since A1 intersect A2 has only two possibilities: either it has A3 or it doesn't, we can say:
P(A3' intersect (A1 intersect A2)) = P(A1 intersect A2) - P(A1 intersect A2 intersect A3) = 0.12 - 0.01 = 0.11.
Now we can compute: P(A3' | A1 intersect A2) = P(A3' intersect (A1 intersect A2)) / P(A1 intersect A2) = 0.11 / 0.12 ≈ 0.917.
For part (b), you are correct that you need to find the probability of the system having all three types of defects given that it has a type 1 defect. To do this, you can use conditional probability.
Let A1 denote the event that the system has a type 1 defect.
Let A1A2A3 denote the event that the system has all three types of defects.
We want to find P(A1A2A3 | A1), which means the probability of all three defects occurring given that the system has a type 1 defect.
Using the definition of conditional probability, we have:
P(A1A2A3 | A1) = P(A1A2A3 and A1) / P(A1)
Now, we know that the system having all three types of defects is equivalent to the intersection of A1, A2, and A3, so:
P(A1A2A3 and A1) = P(A1 intersect A2 intersect A3)
Plugging in the given probability values we have:
P(A1A2A3 | A1) = P(A1 intersect A2 intersect A3) / P(A1)
Thus, you are correct in calculating 0.01 / 0.12 to find the answer to part (b).
For part (c), you need to find the probability that the system has exactly one type of defect given that it has at least one type of defect.
To solve this problem, you can use the concept of complementary probability. The complementary event to having exactly one type of defect is having more than one type of defect or having all three types of defects.
Let E1 denote the event that the system has exactly one type of defect, and let E1' denote the event that the system has more than one type of defect or all three types of defects.
We want to find P(E1 | (A1 union A2 union A3)), which means the probability of having exactly one type of defect given that the system has at least one type of defect.
Using Bayes' theorem, we have:
P(E1 | (A1 union A2 union A3)) = P((A1 union A2 union A3) | E1) * P(E1) / P((A1 union A2 union A3))
Now, we know that (A1 union A2 union A3) is equivalent to the union of events E1 and E1', so:
P((A1 union A2 union A3) | E1) = P(E1)
Plugging in the given probability values we have:
P(E1 | (A1 union A2 union A3)) = P(E1) / P((A1 union A2 union A3))
To find P(E1), you can add up the probabilities of having exactly one defect of each type:
P(E1) = P(A1) * (1 - P(A2 union A3)) + P(A2) * (1 - P(A1 union A3)) + P(A3) * (1 - P(A1 union A2))
To find P((A1 union A2 union A3)), you can use the inclusion-exclusion principle:
P((A1 union A2 union A3)) = P(A1) + P(A2) + P(A3) - P(A1 intersect A2) - P(A1 intersect A3) - P(A2 intersect A3) + P(A1 intersect A2 intersect A3)
With these formulas, you can calculate the answer to part (c).
For part (d), you want to find the probability that the system does not have the third type of defect given that it has both the first two types of defects.
Let F denote the event that the system does not have the third type of defect.
Let G denote the event that the system has both the first two types of defects (A1 and A2).
We want to find P(F | G), which means the probability that the system does not have the third type of defect given that it has both the first two types of defects.
Using Bayes' theorem, we have:
P(F | G) = P(G | F) * P(F) / P(G)
Now, we know that G is equivalent to the intersection of events A1 and A2, so:
P(G | F) = P(A1 intersect A2 | F)
Plugging in the given probability values we have:
P(F | G) = P(A1 intersect A2 | F) * P(F) / P(G)
To find P(F), you can calculate 1 - P(A3), which represents the probability of not having the third type of defect.
To find P(G), you can calculate P(A1 intersect A2) directly from the given probability values.
With these formulas, you can calculate the answer to part (d).
(b) Given that the system has a type 1 defect, the probability that it has all three types of defects can be found using conditional probability.
Let's denote:
B1 = Event that the system has a type 1 defect
B2 = Event that the system has a type 2 defect
B3 = Event that the system has a type 3 defect
We need to find P(B1 intersect B2 intersect B3 | B1), which can be calculated using the formula:
P(B1 intersect B2 intersect B3 | B1) = P((B1 intersect B2 intersect B3) intersect B1) / P(B1)
To calculate the numerator, we use P(A1 intersect A2 intersect A3) = 0.01.
To calculate the denominator, we use P(A1) = 0.12.
Therefore,
P(B1 intersect B2 intersect B3 | B1) = P(A1 intersect A2 intersect A3) / P(A1)
= 0.01 / 0.12
= 0.0833 (rounded to four decimal places)
So, the probability that the system has all three types of defects given that it has a type 1 defect is approximately 0.0833.
(c) Given that the system has at least one type of defect, we need to calculate the probability that it has exactly one type of defect.
Let's denote:
C1 = Event that the system has exactly one type of defect (type 1)
C2 = Event that the system has exactly one type of defect (type 2)
C3 = Event that the system has exactly one type of defect (type 3)
We want to find P(C1 union C2 union C3 | (A1 union A2 union A3)).
To calculate the numerator, we need to determine the probability of having exactly one type of defect:
P(C1) = P(A1) - P(A1 intersect A2) - P(A1 intersect A3) + P(A1 intersect A2 intersect A3)
P(C2) = P(A2) - P(A1 intersect A2) - P(A2 intersect A3) + P(A1 intersect A2 intersect A3)
P(C3) = P(A3) - P(A1 intersect A3) - P(A2 intersect A3) + P(A1 intersect A2 intersect A3)
To calculate the denominator, we need to determine the probability of having at least one type of defect:
P((A1 union A2 union A3)) = P(A1) + P(A2) + P(A3) - P(A1 intersect A2) - P(A1 intersect A3) - P(A2 intersect A3) + P(A1 intersect A2 intersect A3)
Once we calculate the values, we can find the probability using conditional probability:
P(C1 union C2 union C3 | (A1 union A2 union A3))
= (P(C1) + P(C2) + P(C3)) / P((A1 union A2 union A3))
(d) Given that the system has both the first two types of defects, we want to calculate the probability that it does not have the third type of defect.
Let's denote:
D = Event that the system does not have the third type of defect
We need to calculate P(D | (A1 intersect A2)). This can be calculated using conditional probability:
P(D | (A1 intersect A2)) = P(D intersect (A1 intersect A2)) / P(A1 intersect A2)
To calculate the numerator, we need to determine the probability of not having the third type of defect and having the first two types of defects:
P(D intersect (A1 intersect A2)) = P(A1 intersect A2) - P(A1 intersect A2 intersect A3)
To calculate the denominator, we have P(A1 intersect A2) = 0.13 (given).
Once we calculate the values, we can find the probability using conditional probability:
P(D | (A1 intersect A2)) = (P(D intersect (A1 intersect A2)) / P(A1 intersect A2))