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 the following probabilities are true.

P(A1) = 0.12 P(A2) = 0.08 P(A3) = 0.05P(A1 U A2) = 0.14 P(A1 n A3) = 0.14P(A2 U A3) = 0.11 P(A1 n A2 n A3) = 0.01
(a) Given that the system has a type 1 defect, what is the probability that it has a type 2 defect? (Round your answer to four decimal places.)

Question

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Respuesta :

mudamoon97 mudamoon97 · Answer 1 · 2020-03-13T10:05:22+01:00

Answer:

Step-by-step explanation:

a) Given that system has atleast one type of defect and its probability = P(A1 U A2 U A3)

P(A₁) = 0.11

P(A₂) = 0.08

P(A₃) = 0.05

P(A₁uA₂) = 0.13, P(A₁uA₃) = 0.13, P(A₂uA₃) = 0.11

P(A₁nA₂nA₃) = 0.01

The probabilities P(A1 ∩ A2), P(A1 ∩ A3 ), P(A2 ∩ A3 ) are calculated as shown below

P(A₁nA₂) = P(A₁) + P(A₂) - P(A₁uA₂) = 0.11 + 0.08 - 0.13 = 0.06

P(A₁nA₃) = P(A₁) + P(A₃) - P(A₁uA₃) = 0.11 + 0.05 - 0.13 = 0.03

P(A₁nA₂) = P(A₂) + P(A₃) - P(A₂uA₃) = 0.08 + 0.05 - 0.11 = 0.02

The probability that at least one defect in the system

P(A₁uA₂uA₃) = P(A₁) + P(A₂) + P(A₃) - P(A₁nA₂) - P(A₂nA₃) - P(A₁nA₃) + P(A₁nA₂nA₃)

= 0.11 + 0.08 + 0.05 - 0.06 - 0.02 - 0.03 + 0.01

= 0.14

AFOKE88 AFOKE88 · Answer 2 · 2020-03-13T10:49:03+01:00

Answer:

Probability is 0.06

Step-by-step explanation:

From the question, we can see that it says the system already has a type 1 defect.

So for it to have a type 2 defect also, it means it will now have type 1 and 2 defect and thus; P(A1 ∩ A2)

And P(A1 ∩ A2) = P(A1) + P(A2) − P(A1 ∪ A2)

And from the question,

P(A1) = 0.12

P(A2) = 0.08

P(A1 ∪ A2) = 0.14

So, P(A1 ∩ A2) = 0.12 + 0.08 - 0.14 = 0.06