An array of 30 LED bulbs is used in an automotive light. The probability that a bulb is defective is 0.001 and defective bulbs occur independently. Determine the following: a. Probability that an automotive light has two or more defective bulbs. b. Expected number of automotive lights to check to obtain one with two or more defective bulbs.

Answer:

a) 0.0005 = 0.05% probability that an automotive light has two or more defective bulbs.

b) 4000 automotive lights.

Step-by-step explanation:

For each LED, there are only two possible outcomes. Either it is defective, or it is not. Defective bulbs occur independently, which 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.

An array of 30 LED bulbs is used in an automotive light. The probability that a bulb is defective is 0.001.

This means that, respectively, [tex]n = 30, p = 0.001[/tex].

a. Probability that an automotive light has two or more defective bulbs.

This is:

[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]

In which

[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]

So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{30,0}.(0.001)^{0}.(0.999)^{30} = 0.9704[/tex]

[tex]P(X = 1) = C_{30,1}.(0.001)^{1}.(0.999)^{29} = 0.0291[/tex]

[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.9704 + 0.0291 = 0.9995[/tex]

[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.9995 = 0.0005[/tex]

0.0005 = 0.05% probability that an automotive light has two or more defective bulbs.

b. Expected number of automotive lights to check to obtain one with two or more defective bulbs.

0.0005 = 0.05% probability that an automotive light has two or more defective bulbs.

The number of expected trials to obtain n successes with p probability is given by:

[tex]E = \frac{n}{p}[/tex]

In this case, we have that [tex]n = 2, p = 0.0005[/tex]. So

[tex]E = \frac{n}{p} = \frac{2}{0.0005} = 4000[/tex]

4000 automotive lights.

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