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Sagot :
Answer:
a) 0.8802 = 88.02% probability that the shipment is accepted
b) 0.1756 = 17.56% probability that the shipment is accepted if 15% of the total shipment is defective.
Step-by-step explanation:
For each item, there are only two possible outcomes. Either it is defective, or it is not. The probability of an item being defective is independent of any other item. 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.
20 randomly selected video streaming devices
This means that [tex]n = 20[/tex]
a. Find the probability that the shipment is accepted if 3% of the total shipment is defective.
3% defective means that [tex]p = 0.03[/tex]
Probability of at most 1 defective is:
[tex]P(X \leq 1) = P(X = 0) + P(X = 1)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{20,0}.(0.03)^{0}.(0.97)^{20} = 0.5438[/tex]
[tex]P(X = 1) = C_{20,1}.(0.03)^{1}.(0.97)^{19} = 0.3364[/tex]
[tex]P(X \leq 1) = P(X = 0) + P(X = 1) = 0.5438 + 0.3364 = 0.8802[/tex]
0.8802 = 88.02% probability that the shipment is accepted.
b. Find the probability that the shipment is accepted if 15% of the total shipment is defective
Now [tex]p = 0.15[/tex]. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{20,0}.(0.15)^{0}.(0.85)^{20} = 0.0388[/tex]
[tex]P(X = 1) = C_{20,1}.(0.15)^{1}.(0.85)^{19} = 0.1368[/tex]
[tex]P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0388 + 0.1368 = 0.1756[/tex]
0.1756 = 17.56% probability that the shipment is accepted if 15% of the total shipment is defective.
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