# Exponential distribution — calculating probability that computer works!

If X is the number of years a computer works, and it follows an exponential distribution with a lambda 3, what's the probability a computer will work in 8 years?

I'm not sure that I'm going about this the right way and would love to get feedback. I'm following this formula: $P(X \leq x) = 1 - e^{-\lambda x}$ When plugging in values for x=8 and lambda=3, I get that $P(X\leq 8 ) = .9999$...; and to get P(X>8) I subtracted this from 1. Final answer .000000000378

Is this correct? It seems like a very small probability and not sure that I'm going about it the right way. Any ideas are greatly appreciated. Thank you!

## 1 Answer

Seems that the approach is right. I haven't checked your final computations. Keep in mind that since the rate $\lambda$=3 the mean lifetime is 1/3 years. So lasting 8 years is a very long time and therefore P(X>8) should be very small. Note that 8 years is 24 average lifetimes.

• A mean lifetime of 1/3 years is so unrealistic for most computers that we ought to use common sense (as employed admirably by the OP) and interpret "lambda 3" as being a mean of 3 years. – whuber Jan 2 '18 at 1:37
• We don't know what the OP thought. It was the rate parameter as stated on the question. I agree that very few computers breakdown in 1/3 of year. – Michael Chernick Jan 2 '18 at 1:40
• Michael, the word "rate" appears nowhere in the question. I agree that the OP was using "lambda" as if it were a rate--but isn't that error precisely the issue that needs to be addressed, rather than being compounded by repeating it? – whuber Jan 2 '18 at 13:20
• I don't understand what you're referring to. "Too small" compared to what? Regardless, between the two competing interpretations of $\lambda$ only one is plausible. – whuber Jan 2 '18 at 17:19
• "Low" implies a comparison: it has no absolute meaning. The value, by the way, is close to 7%. – whuber Jan 2 '18 at 20:37