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I have a machine part that have lifetime uniformly distributed between 0 years and 1 year. Whenever a part fails, it is immediately replaced with a new identical part. I know that lifetimes of replacement parts are independent.

My question is: what is the probability that 5 years from now, I replace the part exactly 8 times.

I know that the expected lifetime of a bulb is 1/2 years, but I am not sure how to incorporate that into calculating the probability.

One of the approaches I am thinking of is to model the question into a Markov Chain with two states, Good and Bad. I know that P(Bad, Good) = 1 but I am not sure what the transition probabilities starting from state Good would look like?

Maybe I am complicating the problem too much, any help would be appreciated.

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Look at the probability that 8 parts lives sum to less than five years. So take 8 random parts and let $x_1 ... x_8$ be their lifetimes, find $P(x_1 + ... + x_8 \le 5)$ then you need to subtract the probability that 9 sum to less than 5.

To calculate this you need to use a normal approximation since the sum of uniform distributions converge quickly to normal. The mean of the sum is 4 and the variance is $\frac{8}{12}$, so $P(x_1 + ... + x_8 \le 5) = P(Z \le \sqrt{\frac{12}{8}})$. You need to then do the same steps to subtract $P(x_1 + ... + x_9\le 5)$

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I am sorry, but I still don't get it. Could someone explain it at least for a basic case (for example, what it the probability that 3 years from now, I will replace the bulb 4 times)? –  Solver Feb 24 '13 at 15:38
    
@Solver In practice it will be almost impossible to calculate exactly. If you want the exact formula its here: mathworld.wolfram.com/UniformSumDistribution.html. However the sum of normal distribution converges VERY QUICKLY to the normal distribution so I would just do a normal approximation. The mean being the sum of the means and the variance being the sum of the variances. So the mean of the sum of 8 is 4 and the variance 8/12. –  Sean Ballentine Feb 24 '13 at 17:28
    
Ill edit my post. –  Sean Ballentine Feb 24 '13 at 17:28
    
Thanks, this is much clearer. Just one more question to make sure I understand the concept correctly, if I want to calculate the probability that 5 years from now, I will replace the part at least 8 times, then I will not subtract the $P(x_1+...+x_9≤5)$. Is that correct? and thanks again for your help. –  Solver Feb 24 '13 at 17:39
    
@solver Yes, thats exactly right. –  Sean Ballentine Feb 24 '13 at 17:39
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