In a chemical plant, 24 holding tanks are used for final product storage. Four tanks are selected at random and without replacement. Suppose that six of the tanks contain material in which the viscosity exceeds the customer requirements. a. What is the probability that exactly one tank in the sample contains high-viscosity material?
b. What is the probability that at least one tank in the sample contains high-viscosity material?
c. In addition to the six tanks with high viscosity levels, four different tanks contain material with high impurities. What is the probability that exactly one tank in the sample contains high-viscosity material and exactly one tank in the sample contains material with high impurities?

Answer:

a) P(A) = 0,4607     or   P(A) = 46,07 %

b) P(B) = 0,7120   or 71,2 %

c) P(C) = 0,2055  or P(C) = 20,55 %

Step-by-step explanation:

We will use two concepts in solving this problem.

1.- The probability of an event (A) is for definition:

P(A) = Number of favorable events/ Total number of events FE/TE

2.- If A and B are complementary events ( the sum of them is equal to 1) then:

P(A) = 1 - P(B)

a) The total number of events is:

C ( 24,4) = 24! / 4! ( 24 - 4 )!    ⇒  C ( 24,4) = 24! / 4! * 20!

C ( 24,4) = 24*23*22*21*20! / 4! * 20!  

C ( 24,4) = 24*23*22*21/4*3*2

C ( 24,4) = 24*23*22*21/4*3*2    ⇒  C ( 24,4) =  10626

TE = 10626

Splitting the group of tanks in two 6 with h-v  and 24-6 (18) without h-v

we get that total number of favorable events is the product of:

FE = 6* C ( 18, 3)  = 6 * 18! / 3!*15!  =  18*17*16*15!/15!

FE =  4896

Then P(A) ( 1 tank in the sample contains h-v material is:

P(A) = 4896/10626

P(A) = 0,4607     or   P(A) = 46,07 %

b) P(B) will be the probability of at least 1 tank contains h-v

P(B) = 1 - P ( no one tank with h-v)

Again Total number of events is 10626

The total number of favorable events for the ocurrence of P is C (18,4)

FE = C (18,4) = 18! / 14!*4! = 18*17*16*15*14!/14!*4!

FE = 18*17*16*15/4*3*2  = 3060

Then P = 3060/10626

P = 0,2879

And the probability we are looking for is

P(B) = 1 - 0,2879

P(B) = 0,7120   or 71,2 %

c) We call P(C) the probability of finding exactly 1 tank with h-v and t-i

having 4 with t-i tanks is:

reasoning the same way but now having 4 with t-i (impurities) number of favorable events is:

FE = 6*4* C(14,2) = 24 * 14!/12!*2!

FE = 24* 14*13*12! / 12!*2

FE = 24*14*13/2    ⇒  FE = 2184

And again as the TE = 10626

P(C) = 2184/ 10626

P(C) = 0,2055  or P(C) = 20,55 %

Final answer:

These problems relate to the field of probability and specifically utilize the Hypergeometric Distribution. By plugging data into the appropriate formula, we can find the probabilities. For instance, the joint probability of independent events can be used to find the chance of exactly one tank having high viscosity and one tank having high impurities.

Explanation:

This question is asking us to solve probability problems. It specifically related to the Hypergeometric Distribution, which is used when we're interested in success/failure outcomes (in this case, tanks with high or acceptable viscosity), and when we're sampling without replacement.

For part a, we are looking for the probability of picking exactly one tank with high-viscosity. We would calculate this using the hypergeometric distribution formula:

P(X=k) = (C(K, k) * C(N-K, n-k)) / C(N, n)

where K is the total number of success states in the population (6 tanks with high viscosity), k is the number of success states in the sample (1 tank), N is the population size (24 tanks), and n is the number of samples (4 tanks). Plugging these numbers in, we can find the answer.

For part b, to find the probability that at least one tank in the sample contains high-viscosity material, we can either sum the probabilities P(X=1), P(X=2), P(X=3), and P(X=4), or find the complement of the probability that none of the tanks have high viscosity, i.e., 1- P(X=0).

For part c, with the addition of 4 tanks with high impurities, we can use the joint probability of independent events, which is the product of the probabilities of the two independent events. Here, the probability that exactly one tank has high viscosity and exactly one tank has high impurities would be the product of the two individual probabilities calculated in a similar manner.

Learn more about Hypergeometric Distribution here:

brainly.com/question/35126689

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