Question

With one method of a procedure called acceptance sampling, a sample of items is randomly selected without replacement and the entire batch is accepted if every item in the sample is okay. The ABC Electronics Company has just manufactured 1650 write-rewrite CDs, and 100 are defective. If 4 of these CDs are randomly selected for testing, what is the probability that the entire batch will be accepted?

Answer 1

Answer:

77.86% probability that the entire batch will be accepted

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

We also have that the order in which the CD's are chosen is not important. So we use the combinations formula to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

Desired outcomes:

4 CD's from a set of 1550(non defective). So

$D = C_{1550,4} = \frac{1550!}{4!(1550 - 4)!} = 239570392425$

Total outcomes:

4 CD's from a set of 1650. So

$D = C_{1650,4} = \frac{1650!}{4!(1650 - 4)!} = 307711809900$

Probability

$p = \frac{D}{T} = \frac{239570392425}{307711809900} = 0.7786$

77.86% probability that the entire batch will be accepted

Answer 2

Answer:

The probability that the entire batch will be accepted is 0.7774.

Step-by-step explanation:

We are given that with one method of a procedure called acceptance sampling, a sample of items is randomly selected without replacement and the entire batch is accepted if every item in the sample is okay. The ABC Electronics Company has just manufactured 1650 write-rewrite CD's, and 100 are defective.

Also, 4 of these CD's are randomly selected for testing.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 4 CD's

             r = number of success = all 4

   p = probability of success which in our question is probability of

          CD's being not defective in total of 1650 manufactured, i.e;  1 - $\frac{100}{1650}$ =

           1 - 0.061 = 0.939

LET X = Number of CD's that are not defective

So, it means X ~ $Binom(n=4, p=0.939)$

Here, the batch will be accepted when all four selected CD"s are not defective.

Now, Probability that the entire batch will be accepted is given by = P(X = 4)

          P(X = 4) = $\binom{4}{4}\times 0.939^{4}\times (1-0.939)^{4-4}$

                        = $1 \times 0.939^{4} \times 1$

                        = 0.7774

Therefore, Probability that the entire batch will be accepted is 0.7774.