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Sagot :
Answer:
A. 0.33496 = 33.496% probability that one out of the next four cars needs oil.
B. 0.19635 = 19.635% probability that two out of the next eight cars needs oil.
C. 0.01437 = 1.437% probability that 10 out of the next 40 cars needs oil.
Step-by-step explanation:
For each customer, there are only two possible outcomes. Either their car needs to have oil added, or it does not. The probability of a customer having a car needing oil addition is independent of any other customer. 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.
One out of 8 cars needs to have oil added.
This means that [tex]p = \frac{1}{8} = 0.125[/tex]
A. One out of the next four cars needs oil.
This is [tex]P(X = 1)[/tex] when [tex]n = 4[/tex]. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 1) = C_{4,1}.(0.125)^{1}.(0.875)^{3} = 0.33496[/tex]
0.33496 = 33.496% probability that one out of the next four cars needs oil.
B. Two out of the next eight cars needs oil.
This is [tex]P(X = 2)[/tex] when [tex]n = 8[/tex]. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{8,2}.(0.125)^{2}.(0.875)^{6} = 0.19635[/tex]
0.19635 = 19.635% probability that two out of the next eight cars needs oil.
C. 10 out of the next 40 cars needs oil.
This is [tex]P(X = 10)[/tex] when [tex]n = 40[/tex]. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 10) = C_{40,10}.(0.125)^{10}.(0.875)^{30} = 0.01437[/tex]
0.01437 = 1.437% probability that 10 out of the next 40 cars needs oil.
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