You purchased a five pack of new light bulbs that were recalled because 6% of the lights did not work. what is the probability that at least one of your lights is defective

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Answer:

26.61% probability that at least one of your lights is defective

Step-by-step explanation:

For each light, there are only two possible outcomes. Either it works, or it does not. The probability of a light working is independent of other lights. So 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.

You purchased a five pack of new light bulbs that were recalled because 6% of the lights did not work.

Five bulbs, so n = 5

6% do not work, so p = 0.06.

What is the probability that at least one of your lights is defective

Either none is defective, or at least one is. The sum of the probabilities of these events is decimal 1. So

[tex]P(X = 0) + P(X \geq 1) = 1[/tex]

We want [tex]P(X \geq 1)[/tex]. So

[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]

In which

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

[tex]P(X = 0) = C_{5,0}.(0.06)^{0}.(0.94)^{5} = 0.7339[/tex]

[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.7339 = 0.2661[/tex]

26.61% probability that at least one of your lights is defective

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