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Sagot :
Answer:
0.271 = 27.1% probability that at least one will have a defect
Step-by-step explanation:
For each item, there are only two possible outcomes. Either they have a defect, or they do not. Items are independent. This means that we use the binomial probability distribution 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.
A manufacturing machine has a 10% defect rate.
This means that [tex]p = 0.1[/tex]
3 items are chosen at random
This means that [tex]n = 3[/tex]
What is the probability that at least one will have a defect?
This is
[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{3,0}.(0.1)^{0}.(0.9)^{3} = 0.729[/tex]
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.729 = 0.271[/tex]
0.271 = 27.1% probability that at least one will have a defect
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