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 ? A2) = 0.14 P(A1 ? A3) = 0.14P(A2 ? 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.)(b) Given that the system has a type 1 defect, what is the probability that it has all three types of defects? (Round your answer to four decimal places.)(c) Given that the system has at least one type of defect, what is the probability that it has exactly one type of defect? (Round your answer to four decimal places.)(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? (Round your answer to four decimal places.)

(a) 0.5000

(b) 0.0833

(c) 0.2500

((d) 0.9500

Step-by-step explanation:

Given that the following probabilities;

P(A1) = 0.12; P(A2) = 0.08,

P(A3) = 0.05; P(A1 ∪ A2) = 0.14, P(A1 ∪ A3) = 0.14,

P(A2 ∪ A3) = 0.11,

P(A1 ∩ A2 ∩ A3) = 0.01

(a) we want to find P(A2/A1).

P(A2/A1) = P(A1nA2)/P(A1)a

So we need to find P(A1nA2)

P(A1 ∪ A2)= P(A1) + P(A2) - P(A1nA2)

P(A1nA2) =P(A1) + P(A2) - P(A1 ∪ A2)

P(A1nA2) = 0.12 + 0.08 - 0.14

P(A1nA2) = 0.20- 0.14

P(A1nA2 )= 0.060

P(A2/A1) = 0.06/0.12

= 0.5000. (Ans)

(b) The required probability is: P(A2∩A3/A1) =[P(A2∩A3)∩P(A1)] /P(A1)b

But [P(A2nA3)nP(A1)] = P(A1∩A2∩ A3)

P(A2∩A3/A1) = 0.01/0.12

P(A2∩A3/A1) = 0.0833 (Ans)

(c) the required probability is given by Pr =

(Pr. of A1 defect) + (Pr. of A2 defect) + (Pr. of A3 defect)c

Pr = 0.12 + 0.08 + 0.05

Pr = 0.2500 (And)

(d) The required probability is given by:

Pr[(A3)' /(A1∩A2)] = Pr[(A3)'nPr(A1∩A2)] /Pr(A1∩A2)]

Pr(A3)’ = 1 - Pr(A3) = 1 -0.05

Pr(A3)'/(A1∩A2)= (0.95×0.06)/(0.06)

Pr(A3)'/(A1∩A2 )= 0.95(Ans)d

2,600 divided by 80 = 32.5

step-by-step explanation:

when rotating a figure 180 degrees, imagine that you are able to take the figure and turn it clockwise or counterclockwise around a center point. to rotate a figure 180 degrees, you will need to apply the rule (x, y) → (-x, -y). start by using a coordinate grid with coordinates for each vertex of the figure. the center point of the coordinate grid is located at (0, 0), which is what you will rotate the figure around. write down the original coordinates of the shape you are going to rotate. then, apply the rule. for example, coordinate (1, 2) becomes (-1, -2). another example would be, coordinate (-4, -2) becoming (4, 2). after you change each original coordinate to the rotated coordinate, you will draw your new figure.

step-by-step explanation: