The source of the error when a computer system fails are either the disk drive, the computer memory, or the operating system. Very often, 50% of the errors are disk drive errors, 20% are computer memory errors, and the remainder are operating system errors. From the component performance standards, the probabilities of failure due to disk drive, computer memory, and operating system errors are 0.4, 0.6 and 0.25, respectively. Given the information from the component performance standards, what is the probability of an operating system error, given that a failure occurred

The probability of an operating system error given that a failure has ocurred is P(O/E)= 0.19

Step-by-step explanation:

Hello!

There are three different sources of error when a computer system fails:

D: disk drive error. ⇒ P(D)= 0.50

M: computer memory error. ⇒P(M)= 0.20

O: operating systems error. ⇒ P(O)= 0.30

According to the component perdormance standard:

The probability of failure "E", given that there is a disk drive error is P(E/D)= 0.40

The probability of failure "E", given that there is a computer memory error is P(E/M)= 0.6

The probability of failure "E", given that there is a operating system error is P(E/O)= 0.25

You need to calculate the probability of an operating system error given that a failure has ocurred, symbolically:

P(O/E)

To reach the probability of the marginal E you have to add all intersections between the event "E" and the events "D", "M" and "O"

       D           M             O       Total

E:  P(E∩D); P(E∩M); P(E∩O);  P(E)

Using the formula of the conditional probability I'll clear all three intersections using the known probabilities:

General formula: ⇒

P(E∩D)= P(E/D)*P(D)= 0.4*0.5= 0.2

P(E∩M)= P(E/M)*P(M)= 0.6*0.2= 0.12

P(E∩O)= P(E/O)*P(O)= 0.25*0.3= 0.075

P(E)= P(E∩D) + P(E∩M) + P(E∩O)= 0.2+0.12+0.075= 0.395

I hope it helps!

function 2 has a greater rate of change than function 1

function 2 has a greater y-intercept than function 1

step-by-step explanation: