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Sagot :
Using the binomial distribution, it is found that:
0.3677 = 36.77% probability that more than 1 of these parts are defective.
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For each part, there are only two possible outcomes. Either they are defective, or they are not. The probability of a part being defective is independent of any other part, which 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 a success.
At a certain auto parts manufacturer, the Quality Control division has determined that one of the machines produces defective parts 18% of the time.
This means that [tex]p = 0.18[/tex]
A random sample of 7 parts produced by this machine is chosen.
This means that [tex]n = 7[/tex]
Find the probability that more than 1 of these parts are defective.
This is:
[tex]P(X \geq 1) = 1 - P(X < 1)[/tex]
In which
[tex]P(X < 1) = P(X = 0) + P(X = 1)[/tex]
Thus
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{7,0}.(0.18)^{0}.(0.82)^{7} = 0.2493[/tex]
[tex]P(X = 1) = C_{7,1}.(0.18)^{1}.(0.82)^{6} = 0.3830[/tex]
Then
[tex]P(X < 1) = P(X = 0) + P(X = 1) = 0.2493 + 0.3830 = 0.6323[/tex]
[tex]P(X \geq 1) = 1 - P(X < 1) = 1 - 0.6323 = 0.3677[/tex]
0.3677 = 36.77% probability that more than 1 of these parts are defective.
For more on the binomial probability distribution, you can check https://brainly.com/question/15557838
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