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Sagot :
Answer:
a) 0.0645 = 6.45% probability that the 90 orders can be filled without reordering components.
b) 0.4062 = 40.62% probability that the 100 orders can be filled without reordering components.
c) 0.9034 = 90.34% probability that the 100 orders can be filled without reordering components
Step-by-step explanation:
For each component, there are only two possible outcomes. Either it is defective, or it is not. The components can be assumed to be 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.
3% of the components are identified as defective
This means that [tex]p = 0.03[/tex]
a. If the manufacturer stocks 90 components, what is the probability that the 90 orders can be filled without reordering components?
0 defective in a set of 90, which is [tex]P(X = 0)[/tex] when [tex]n = 90[/tex]. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{90,0}.(0.03)^{0}.(0.97)^{90} = 0.0645[/tex]
0.0645 = 6.45% probability that the 90 orders can be filled without reordering components.
b. If the manufacturer stocks 102 components, what is the probability that the 100 orders can be filled without reordering components?
At most 102 - 100 = 2 defective in a set of 102, so [tex]P(X \leq 2)[/tex] when [tex]n = 102[/tex]
Then
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{102,0}.(0.03)^{0}.(0.97)^{102} = 0.0447[/tex]
[tex]P(X = 1) = C_{102,0}.(0.03)^{1}.(0.97)^{101} = 0.1411[/tex]
[tex]P(X = 2) = C_{102,2}.(0.03)^{2}.(0.97)^{100} = 0.2204[/tex]
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0447 + 0.1411 + 0.2204 = 0.4062[/tex]
0.4062 = 40.62% probability that the 100 orders can be filled without reordering components.
c. If the manufacturer stocks 105 components, what is the probability that the 100 orders can be filled without reordering components?
At most 105 - 100 = 5 defective in a set of 105, so [tex]P(X \leq 5)[/tex] when [tex]n = 105[/tex]
Then
[tex]P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{105,0}.(0.03)^{0}.(0.97)^{105} = 0.0408[/tex]
[tex]P(X = 1) = C_{105,0}.(0.03)^{1}.(0.97)^{104} = 0.1326[/tex]
[tex]P(X = 2) = C_{105,2}.(0.03)^{2}.(0.97)^{103} = 0.2133[/tex]
[tex]P(X = 3) = C_{105,3}.(0.03)^{3}.(0.97)^{102} = 0.2265[/tex]
[tex]P(X = 4) = C_{105,4}.(0.03)^{4}.(0.97)^{101} = 0.1786[/tex]
[tex]P(X = 5) = C_{105,5}.(0.03)^{5}.(0.97)^{100} = 0.1116[/tex]
[tex]P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0408 + 0.1326 + 0.2133 + 0.2265 + 0.1786 + 0.1116 = 0.9034[/tex]
0.9034 = 90.34% probability that the 100 orders can be filled without reordering components
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