Respuesta :
Answer:
a) There is a 59.87% probability that none of the LED light bulbs are defective.
b) There is a 31.51% probability that exactly one of the light bulbs is defective.
c) There is a 98.84% probability that two or fewer of the LED light bulbs are defective.
d) There is a 100% probability that three or more of the LED light bulbs are not defective.
Step-by-step explanation:
For each light bulb, there are only two possible outcomes. Either it fails, or it does not. This means that we use the binomial probability distribution to solve this problem.
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 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 X happening.
In this problem we have that:
[tex]n = 10, p = 0.05[/tex]
a) None of the LED light bulbs are defective?
This is P(X = 0).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{10,0}*(0.05)^{0}*(0.95)^{10} = 0.5987[/tex]
There is a 59.87% probability that none of the LED light bulbs are defective.
b) Exactly one of the LED light bulbs is defective?
This is P(X = 1).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 1) = C_{10,1}*(0.05)^{1}*(0.95)^{9} = 0.3151[/tex]
There is a 31.51% probability that exactly one of the light bulbs is defective.
c) Two or fewer of the LED light bulbs are defective?
This is
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
[tex]P(X = 2) = C_{10,2}*(0.05)^{2}*(0.95)^{8} = 0.0746[/tex]
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.5987 + 0.3151 + 0.0746 0.9884[/tex]
There is a 98.84% probability that two or fewer of the LED light bulbs are defective.
d) Three or more of the LED light bulbs are not defective?
Now we use p = 0.95.
Either two or fewer are not defective, or three or more are not defective. The sum of these probabilities is decimal 1.
So
[tex]P(X \leq 2) + P(X \geq 3) = 1[/tex]
[tex]P(X \geq 3) = 1 - P(X \leq 2)[/tex]
In which
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
[tex]P(X = 0) = C_{10,0}*(0.95)^{0}*(0.05)^{10}\cong 0[/tex]
[tex]P(X = 1) = C_{10,1}*(0.95)^{1}*(0.05)^{9} \cong 0[/tex]
[tex]P(X = 2) = C_{10,1}*(0.95)^{2}*(0.05)^{8} \cong 0[/tex]
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0[/tex]
[tex]P(X \geq 3) = 1 - P(X \leq 2) = 1[/tex]
There is a 100% probability that three or more of the LED light bulbs are not defective.
