Respuesta :
Answer:
0.4546
Step-by-step explanation:
nCr = n!/(n-r)!r!
Number of ways of selecting the four defective voltage regulators from 12 = 12C4 = 12!/(12-4)!4! = 12!/8!4! = (12 *11*10*9)/(4*3*2*1)
12C4 = 495 ways
Number of ways of selecting 2 defectives from line 1 = 6C2 * 6C2
6C2 = 6!/(6-2)!2! = 6!/4!2! = (6*5)/(2*1) = 15
6C2 * 6C2 = 15*15 = 225 ways
Probability = Number of possible outcomes/ Number of total outcomes
Probability that exactly 2 of the defective regulators came from line 1 = 225/40.95 = 0.4546
Using the hypergeometric distribution, it is found that there is a 0.0909 = 9.09% probability that exactly two of the defective regulators came from line I.
The regulators are chosen without replacement, hence the hypergeometric distribution is used to solve this question.
What is the hypergeometric distribution formula?
The formula is:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
- x is the number of successes.
- N is the size of the population.
- n is the size of the sample.
- k is the total number of desired outcomes.
In this problem:
- There are 12 regulators, hence N = 12.
- Four regulators are defective, two from each line, hence k = 2.
- A total of four defective regulators was there, hence n = 4.
The probability is P(X = 2), hence:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]P(X = 2) = h(2,12,4,2) = \frac{C_{2,2}C_{10,2}}{C_{12,4}} = 0.0909[/tex]
0.0909 = 9.09% probability that exactly two of the defective regulators came from line I.
More can be learned about the hypergeometric distribution at https://brainly.com/question/24826394
