The life times, Y in years of a certain brand of low-grade lightbulbs follow an
ID: 3255616 • Letter: T
Question
The life times, Y in years of a certain brand of low-grade lightbulbs follow an exponential distribution with a mean of 0.65 years. A tester makes random observations of the life times of this particular brand of lightbulbs and records them one by one as either a success if the life time exceeds 1 year, or as a failure otherwise. Find the probability to 3 decimal places that the first success occurs in the fifth observation. Find the probability to 3 decimal places of the second success occurring in the 8th observation given that the first success occurred in the 3rd observation. Find the probability to 2 decimal places that the first success occurs in an odd-numbered observation. That is, the first success occurs in the 1st or 3rd or 5th or 7th (and so on) observation.Explanation / Answer
lambda = 1/0.65 = 1.5385
here x = 1Probability that bulb will last more than 1 year is given by, p = e^(-lambda * x)
p = e^(-1.5385)
p = 0.2147
(A) As the further process is binomial i.e. the tester is checking bulb one by one and the outcome is either the bulb will last for more not 1 year or not. This means it is yes or no type of outcome.
The probability that the first success occurs in the fifth observation = (1-p)^4 * p = (1 - 0.2147)^4 * 0.2147 = 0.0817
(B)
P(A|B) = P(A and B) / P(B)
Here P(B) is the probability that first success occurs in the 3rd observation.
P(A|B) is the required probability that the second success occurs in the 8th observation given that first success occurs in 3rd observation
P(A and B) is the probability that the first success occurs in 3rd observation and second success occurs in 8th observation.
P(B) = (1-p)^2 * p = (1 - 0.2147)^2 * 0.2147 = 0.1324
P(A and B) = (1-p)^2 * p * (1-p)^4 * p = (1 - 0.2147)^2 * 0.2147* (1 - 0.2147)^4 * 0.2147 = 0.0108
P(A|B) = 0.0108 / 0.1324 = 0.0815
(C)
sum of these probability will be equal to 0.5484 i.e. 0.54
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