Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If a satellite has two essential parts ( A and B) and the satellite only works if both parts are working. If both parts have a independent, random and exponential usage life; A with parameter 2 years and B parameter 1 year.

a. What is the probability that the satellite will not work a year from launched?

b. what is the probability that the satellite will not work a year from launched and that A is the cause of failure.

Its from stochastic system course, part of Markov unit.

answer:

a)

A~Exp(2) B~Exp(1)

P(A≤1)P(B≤1)=∫(0 to 1) 2e^(-2t) dt ∫(0 to 1) e^(-t) dt = 0,547

∫(0 to 1) 2e^(-2t) dt ∫(1 to ∞) e^(-t) dt = 0,318

∫(0 to 1) e^(-t) dt ∫(1 to ∞) 2e^(-2t) dt = 0,085

P(A≤1)+P(B≤1)+P(A≤1)P(B≤1)=0.95

b) P(fail / A fail) = 0.547/(0.547+0.318)=0.6323

share|improve this question
4  
This seems like homework to me. If it is, then (A) please read this meta post to learn how to ask a homework question; and (B) add the [homework] tag. (If it's not homework, my apologies; just ignore this comment.) –  Srivatsan Jan 4 '12 at 2:41
1  
Your answer to part a) is incorrect. You need to find $P(\{A \leq 1\}\cup\{B\leq 1\})$ and this is not equal to $P\{A \leq 1\}+P(\{B\leq 1\})+P(\{A \leq 1\})P(\{B\leq 1\})$ –  Dilip Sarwate Jan 4 '12 at 3:41
1  
If $T$ has density function $\lambda e^{-\lambda t}$ (for $t \ge 0$), then $\lambda$ is called the parameter of the distribution. The mean of $T$ is then $\frac{1}{\lambda}$ (say hours). The units of the parameter are hours$^{-1}$, not hours. Your statement that the parameter is $2$ hours is puzzling, and non-standard. Probably your book or instructor did not write in this way. Maybe they wrote mean is $2$ hours, meaning that $\lambda=1/2$. Maybe they wrote the parameter is $2$, in which case $\lambda=2$. Please clarify. Once you do, and by tomorrow morning, I can answer the question. –  André Nicolas Jan 4 '12 at 5:45
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.