$Exponential random variables$

Question

$Exponential random variables$

The duration of each solar panel installed in a satellite orbiting the Earth until some failure renders it unusable is modelled by an exponential random variable whose mean is 4 months. If this particular model of satellite is powered by 3 independent solar panels and needs at least 1 of them to function in order to remain operational, what is the probability that the satellite will remain operational for at least 14 months?

Probability

Mathlearner

28

Report

Answer

The answer is accepted.

Join Matchmaticians Affiliate Marketing Program to earn up to a 50% commission on every question that your affiliated users ask or answer.

Answer 1

Answers can only be viewed under the following conditions:

The questioner was satisfied with and accepted the answer, or
The answer was evaluated as being 100% correct by the judge.

View the answer

Since the mean duration of a solar panel is 4 months and the duration is an exponential random variable we know that the density function for failure time (with time given in months) is given by:

$f(x)=\frac{1}{4}e^{-\frac{1}{4}x}$. We can be saved some extra work by knowing the cumulative distribution function as well $F(x)=1-e^{-\frac{1}{4}x}$.

The odds of a given panel failing in 14 months are therefore $F(14)-F(0)$, since $F(0)=0$ the odds of a panel failing are $F(14)$.

The sattellite only fails if the three independent panels fail, so the odds of the sattelite failing in 14 months are $F(14)^{3}$, i.e. each panel fails within 14 months. The probability of the sattelite remaining operational after 14 months is therefore the complement of the probability of the sattelite failing, $1-F(14)^3$ with F defined as earlier.

Poincare

133

$Exponential random variables$

Answer

Related Questions

Search