Answer:
P(T∩E) = 0.017
Step-by-step 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
