Answer: 0.39 ; 0.993
Step-by-step explanation:
Given that:
Mean of exponential distribution = 6
Using poisson :
P(X =x)Probability of failure in next 3 years :
Number of occurrences λ = 1 / 6 = 0.1667
P(X < 3) = 1 - e^-λx
P(X < 3) = 1 - e^-0.1667*3
P(X < 3) = (1 - e^-0.5)
P(X < 3) = 1 - 0.60653
P(X < 3) = 0.39346
= 0.393
B.)
Number of pc owned = 10 ; n = 10
P(atleast 1 fails) :
P(X ≥ 3)^n = 1 - [1 - P(X < 3)] ^n
P(X ≥ 3)^n = 1 - (1 - 0.393)^10
P(X ≥ 3)^n = 1 - 0.607^10
= 1 - 0.00679
= 0.993
This question is based on the probability. Therefore, the probability that the CPU fails in the next three years is 0.393 and the probability that at least one fails within the next three years is 0.993.
Given:
The CPU of a personal computer has a lifetime that is exponentially distributed with a mean lifetime of six years. You have owned this CPU for three years.
(a) We need to determined the probability that the CPU fails in the next three year.
It is given that, mean of exponential distribution = 6
By using Poisson distribution:
P(X =x) Probability of failure in next 3 years :
[tex]Number\, of \,occurrences\, \lambda = \dfrac{1}{6} = 0.1667[/tex]
[tex]P(X < 3) = 1 - e^{-\lambda x}\\P(X < 3) = 1 - e^{-0.1667\times 3}\\P(X < 3) = (1 - e^{-0.5})\\P(X < 3) = 1 - 0.60653\\P(X < 3) = 0.39346= 0.393[/tex]
(b) We need to determined the probability that at least one fails within the next three years.
Number of pc owned = 10 ; n = 10
P(at least 1 fails) :
[tex]P(X \geq 3)^n = 1 - [1 - P(X < 3)] ^n\\P(X \geq 3)^n = 1 - (1 - 0.393)^{10}\\P(X \geq 3)^n = (1 - 0.607)^{10} = = 1 - 0.00679 = 0.993[/tex]
Therefore, the probability that the CPU fails in the next three years is 0.393 and the probability that at least one fails within the next three years is 0.993.
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https://brainly.com/question/11234923
