Respuesta :
Answer:
a) The state space of X is the number of lamps that are defective.
b) Binomial probability mass function.
[tex]P(X = x) = C_{1000,x}.(0.999)^{x}.(0.001)^{1000-x}[/tex]
c) There is a 36.77% probability that none of the lamps are defective.
d) There is a 8.02% probability that more than 2 of the lamps are defective
Step-by-step explanation:
(a) What is the state space of X?
The state space of X is the number of lamps that are defective.
(b) What kind of probability mass function does X have?
X has a binomial probability function mass, because this is the probability of exactly x successes on n repeated trials, and X can only have two outcomes(either the lamp is defective or it is not).
The binomial probability is given by the following formula:
[tex]P(X = x) = C_{n,x}.p^{n}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinatios 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.
In this problem
There are 1000 total lamps, so [tex]n = 1000[/tex].
The probability that any individual lamp is defective is 0.001. A success is a lamp not being defective, so [tex]p = 1 - 0.001 = 0.999[/tex]
The probability of there being x defective pieces is given by the following formula
[tex]P(X = x) = C_{1000,x}.(0.999)^{x}.(0.001)^{1000-x}[/tex]
(c) What is the probability that none of the lamps are defective?
This is [tex]P(X = 1000)[/tex]
[tex]P(X = x) = C_{1000,x}.(0.999)^{x}.(0.001)^{1000-x}[/tex]
[tex]P(X = 1000) = C_{1000,1000}.(0.999)^{1000}.(0.001)^{1000-1000}[/tex]
[tex]P(X = 1000) = (0.999)^{1000} = 0.3677[/tex]
There is a 36.77% probability that none of the lamps are defective.
(d) What is the probability that more than 2 of the lamps are defective?
This is [tex]P(X < 998)[/tex]. The easier way to calculate this is by subtracting 1 by the probability that there is less than 2 defective lamps, by the following formula:
[tex]P(X < 998) = 1 - (P(X = 1000) + P(X = 999) + P(X = 998))[/tex]
[tex]P(X = 1000) = 0.3677[/tex]
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[tex]P(X = x) = C_{1000,x}.(0.999)^{x}.(0.001)^{1000-x}[/tex]
[tex]P(X = 999) = C_{1000,999}.(0.999)^{999}.(0.001)^{1000-999}[/tex]
[tex]P(X = 999) = 1000*(0.999)^{999}*(0.001) = 0.3681[/tex]
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[tex]P(X = x) = C_{1000,x}.(0.999)^{x}.(0.001)^{1000-x}[/tex]
[tex]P(X = 998) = C_{1000,998}.(0.999)^{998}.(0.001)^{1000-998}[/tex]
[tex]P(X = 998) = 499500*(0.999)^{998}*(0.001)^{2} = 0.1840[/tex]
So:
[tex]P(X < 998) = 1 - (P(X = 1000) + P(X = 999) + P(X = 998)) = 1 - (0.3677 + 0.3681 + 0.1840) = 0.0802[/tex]
There is a 8.02% probability that more than 2 of the lamps are defective
