# Exponential Distribution with possible Binomial Probability

You have a system with 6 components. In order for the system to work the following must be met:

Component 1 must work.
At least one of components 2, 3, 4 must work.
At least one of components 5, 6 must work.

Component 1 has an exponentially distributed mean lifetime of 1/2 a year.
Components 2,3,4 have an exponentially distributed mean lifetime of 1 year.
Components 5,6 have an exponentially distributed mean lifetime of 3/2 years.

All components function independently.

What is the probability that the system will function for at least 2 years?

So this is my thinking:

For each group of components, I find the probability that they will last less than 2 years using the cdf of exponential distribution. Then I subtract that probability from 1 to get the probability that they will last at least 2 years. If we define $p_i$ as the probability that the $i^{th}$ component will last 2 years, we have: \begin{align*} p_1 = \exp(-2\times 2) &= 0.018\\ p_2 = p_3 = p_4 = \exp(-1 \times 2) &= 0.135\\ p_5 = p_6 = \exp\{-2/3 \times 2\} &= 0.263. \end{align*} Then, using binomial probability, I find the probability that all of the components will break. Then, I subtract that from 1 to find the probability that at least one of the components will not break. \begin{align*} 1 - b(0;3,0.135) &= 1-(1-0.135)^3 &= 0.354\\ 1 - b(0;2,0.263) &= 1-(1-0.263)^2 &= 0.458 \end{align*} Finally, I multiply the 3 probabilities for the 3 different groups together to find my final answer. $$(0.018)\times(0.354)\times(0.458) = 0.00296$$ So am I right?

• I had written it as a percent, but thank you for verifying. – picokol Sep 23 '15 at 15:57
• OK, my idiosyncrasy: I dislike percents in probability statements... – Xi'an Sep 23 '15 at 17:55
• Is this a homework question? If so, please add the self-study tag. – jlimahaverford Sep 28 '15 at 16:27

$$(1-p_2) \cdot (1-p_3) \cdot (1-p_4).$$