Question

Before the distribution of certain statistical software, every fourth compact disk (CD) is tested for 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) What is the probability that a to-be-tested CD fails any test?
(b) Given that a CD was tested, what is the probability that it failed program 2 or 3?
(c) In a sample of 100, how many CDs would you expect to be rejected?

Answer 1

Answer:

a) For this case we have 4 programs so then if we define the event R that a CD is tested we have the following probability for each test:

$P(R) =\frac{1}{4} =0.25$

The failure probability for each program are given by:

$P(F_1) = 0.01 , P(F_2) = 0.03 , P(F_3) = 0.02 , P(F_4) = 0.01$

For this case we assume that each test is independet form the others.

We can calculate the probability that all 4 programs works properly like this:

$P(4 work) = (1-0.01)*(1-0.03)*(1-0.02)*(1-0.01)= 0.932$

So then the probability that any program fails would be given by:

$P(F) = 1- 0.932= 0.068$

And if we use the fact that we have 4 possible test the true probability of interest would be:

$P(R \cap F) = P(R)*P(F) = 0.25*0.068=0.017$

b) $p= P(F'_1) P(F'_4) *(1- P(F'_2)*P(F'_3))$

And replacing we got:

$p =(1-0.01)*(1-0.01) *[1- (1-0.03)(1-0.02)]= 0.99*0.99*[1- 0.97*0.98]= 0.0484$

c) From part a we now that the probability that any program fails would be given by:

$P(F) = 1- 0.932= 0.068$

So then if we have 100 CDs the expected number of rejected Cd's are:

$100*0.068= 6.8 \approx 7$

Step-by-step explanation:

Part a

For this case we have 4 programs so then if we define the event R that a CD is tested we have the following probability for each test:

$P(R) =\frac{1}{4} =0.25$

The failure probability for each program are given by:

$P(F_1) = 0.01 , P(F_2) = 0.03 , P(F_3) = 0.02 , P(F_4) = 0.01$

For this case we assume that each test is independet form the others.

We can calculate the probability that all 4 programs works properly like this:

$P(4 work) = (1-0.01)*(1-0.03)*(1-0.02)*(1-0.01)= 0.932$

So then the probability that any program fails would be given by:

$P(F) = 1- 0.932= 0.068$

And if we use the fact that we have 4 possible test the true probability of interest would be:

$P(R \cap F) = P(R)*P(F) = 0.25*0.068=0.017$

Part b

For this case we want the probability that it failed program 2 or 3

So then we can find this probability like this:

$p= P(F'_1) P(F'_4) *(1- P(F'_2)*P(F'_3))$

And replacing we got:

$p =(1-0.01)*(1-0.01) *[1- (1-0.03)(1-0.02)]= 0.99*0.99*[1- 0.97*0.98]= 0.0484$

Part c

From part a we now that the probability that any program fails would be given by:

$P(F) = 1- 0.932= 0.068$

So then if we have 100 CDs the expected number of rejected Cd's are:

$100*0.068= 6.8 \approx 7$