Mathematics, 01.10.2019 23:00 offmylawn

Ashop receives a shipment of 1000 cheaply made lamps. the probability that any individual lamp is defective is 0.001. assume the defectiveness is independent for each lamp. let x be the number of defective lamps in the batch of 1000. (a) what is the state space of x? (b) what kind of probability mass function does x have? (c) what is the probability that none of the lamps are defective? (d) what is the probability that more than 2 of the lamps are defective?

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Answer from: ddarby

a) The state space of X is the number of lamps that are defective.

b) Binomial probability mass function.

$P(X = x) = C_{1000,x}.(0.999)^{x}.(0.001)^{1000-x}$

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:

$P(X = x) = C_{n,x}.p^{n}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of a success.

In this problem

There are 1000 total lamps, so n = 1000 .

The probability that any individual lamp is defective is 0.001. A success is a lamp not being defective, so p = 1 - 0.001 = 0.999

The probability of there being x defective pieces is given by the following formula

$P(X = x) = C_{1000,x}.(0.999)^{x}.(0.001)^{1000-x}$

This is P(X = 1000)

$P(X = x) = C_{1000,x}.(0.999)^{x}.(0.001)^{1000-x}$

$P(X = 1000) = C_{1000,1000}.(0.999)^{1000}.(0.001)^{1000-1000}$

$P(X = 1000) = (0.999)^{1000} = 0.3677$

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 P(X < 998) . 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:

P(X < 998) = 1 - (P(X = 1000) + P(X = 999) + P(X = 998))

P(X = 1000) = 0.3677

------------------

$P(X = x) = C_{1000,x}.(0.999)^{x}.(0.001)^{1000-x}$

$P(X = 999) = C_{1000,999}.(0.999)^{999}.(0.001)^{1000-999}$

$P(X = 999) = 1000*(0.999)^{999}*(0.001) = 0.3681$

-------------------

$P(X = x) = C_{1000,x}.(0.999)^{x}.(0.001)^{1000-x}$

$P(X = 998) = C_{1000,998}.(0.999)^{998}.(0.001)^{1000-998}$

$P(X = 998) = 499500*(0.999)^{998}*(0.001)^{2} = 0.1840$

So:

P(X < 998) = 1 - (P(X = 1000) + P(X = 999) + P(X = 998)) = 1 - (0.3677 + 0.3681 + 0.1840) = 0.0802

There is a 8.02% probability that more than 2 of the lamps are defective

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Answer from: Quest

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