Question

Before the distribution of certain statistical software, every fourth compact disk (CD) is testedfor accuracy. The testing process consists of running four independent programs and checking the results. The failure rates for the four testing programs are, respectively, 0.01, 0.03, 0.02, and 0.01.a.(4pts) What is the probability that a CD was tested and failed any test

Answer 1

P(T∩E) = 0.017

Explanation:

Since every fourth CD is tested. Thus if T is the event that represents 4 disks being tested,

P(T) = 1/4 = 0.25

Let Fi represent event of failure rate. So from the question,

P(F1) = 0.01 ; P(F2) = 0.03 ; P(F3) =0.02 ; P(F4) = 0.01

Also Let F'i represent event of success rate. And we have;

P(F'1) = 1 - 0.01 = 0.99 ; P(F'2) = 1 - 0.03 = 0.97; P(F'3) = 1 - 0.02 = 0.98; P(F'4) = 1 - 0.01 = 0.99

Since all programs run independently, the probability that all programs will run successfully is;

P(All programs to run successfully) =

P(F'1) x P(F'2) x P(F'3) x P(F'4) =

0.99 x 0.97 x 0.98 x 0.97 = 0.932

Now, that all 4 programs failed will be = 1 - 0.932 = 0.068

Let E be denote that the CD fails the test. Thus P(E) = 0.068

Now, since testing and CD's defection are independent events, the probability that one CD was tested and failed will be =P(T∩E) = P(T) x P(E)= 0.25 x 0.068 = 0.017