MATH SOLVE

4 months ago

Q:
# The life of a red bulb used in a traffic signal can be modeled using an exponential distribution with an average life of 24 months. The maintenance worker inspects the signal every year. • What is probability that the red bulb will need to be replaced at the first inspection? • If the bulb is in good condition at the end of 18 months, what is the probability that the bulb will be in good condition at the end of 24 months? HINT: Use P(A | B) = P(A∩B) P(B) • If the signal has six red bulbs, what is the probability that atleast one of them needs replacement at the first inspection? Assume distribution of lifetime of each bulb is independent.

Accepted Solution

A:

Answer:See steps belowStep-by-step explanation:Let X be the random variable that measures the lifespan of a bulb.
If the random variable X is exponentially distributed and X has an average value of 24 month, then its probability density function is
[tex] \bf f(x)=\frac{1}{24}e^{-x/24}\;(x\geq 0)[/tex]
and its cumulative distribution function (CDF) is
[tex] \bf P(X\leq t)=\int_{0}^{t} f(x)dx=1-e^{-t/24}[/tex]
• What is probability that the red bulb will need to be replaced at the first inspection?
The probability that the bulb fails the first year is
[tex] \bf P(X\leq 12)=1-e^{-12/24}=1-e^{-0.5}=0.39347[/tex]
• If the bulb is in good condition at the end of 18 months, what is the probability that the bulb will be in good condition at the end of 24 months?
Let A and B be the events,
A = “The bulb will last at least 24 months”
B = “The bulb will last at least 18 months”
We want to find P(A | B).
By definition P(A | B) = P(A∩B)P(B)
but B⊂A, so A∩B = B and [tex] \bf P(A | B) = P(B)P(B) = (P(B))^2[/tex]
We have [tex] \bf P(B)=P(X>18)=1-P(X\leq 18)=1-(1-e^{-18/24})=e^{-3/4}=0.47237[/tex]
hence,
[tex] \bf P(A | B)=(P(B))^2=(0.47237)^2=0.22313[/tex]
• If the signal has six red bulbs, what is the probability that at least one of them needs replacement at the first inspection? Assume distribution of lifetime of each bulb is independent
If the distribution of lifetime of each bulb is independent, then we have here a binomial distribution of six trials with probability of “success” (one bulb needs replacement at the first inspection) p = 0.39347
Now the probability that exactly k bulbs need replacement is
[tex] \bf \binom{6}{k}(0.39347)^k(1-0.39347)^{6-k}[/tex]
Probability that at least one of them needs replacement at the first inspection = 1- probability that none of them needs replacement at the first inspection.
This means that,
Probability that at least one of them needs replacement at the first inspection = [tex] \bf 1-\binom{6}{0}(0.39347)^0(1-0.39347)^{6}=1-(0.60653)^6=0.95021[/tex]