Respuesta :
Answer:
0.148 = 14.8% probability that they will need to order at least one more new transmission
Step-by-step explanation:
For each transmission, there are only two possible outcomes. Either it is defective after a year of use, or it is not. The probability of a transmission being defective is independent of any other transmission. This means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
20% of all the transmissions it installed in a particular style of truck are defective after a year of use.
This means that [tex]p = 0.2[/tex]
Sold seven trucks:
This means that [tex]n = 7[/tex]
It has two of the new transmissions in stock. What is the probability that they will need to order at least one more new transmission?
This is the probability that at least 3 are defective, that is:
[tex]P(X \geq 3) = 1 - P(X < 3)[/tex]
In which
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{7,0}.(0.2)^{0}.(0.8)^{7} = 0.2097[/tex]
[tex]P(X = 1) = C_{7,1}.(0.2)^{1}.(0.8)^{6} = 0.3670[/tex]
[tex]P(X = 2) = C_{7,2}.(0.2)^{2}.(0.8)^{5} = 0.2753[/tex]
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2097 + 0.3670 + 0.2753 = 0.852[/tex]
[tex]P(X \geq 3) = 1 - P(X < 3) = 1 - 0.852 = 0.148[/tex]
0.148 = 14.8% probability that they will need to order at least one more new transmission
