# Probability of success

Suppose I'm building machines that are each made of $n$ identical components, and each of these components has a probability $p$ of being sound (free of faults). Assume that the components are connected "in series", so that the machine is sound only if all its components are sound. Then clearly the probability is $p^n$ that a machine with $n$ components is sound.

Now suppose that I have a population of such machines in which the number of components per machine is normally distributed with a mean of $\mu$ and a standard deviation of $\sigma$.

If I pick a machine at random, what it the probability that it's sound?

If it makes a difference, $p \approx 0.999$ and $n$ is somewhere between a few dozen and a few thousand, in typical situations.

It seems likely that this is a well-known problem in statistics, so just getting a reference or two might be sufficient. I'm interested in computing numerical answers.

• Side note: change "in series" to "sequentially". Commented Feb 9, 2015 at 13:36
• I'd like to see the design of the machine that has $100+\sqrt2$ components. It reminds me of the story about how the average family has $2.3$ children and how tough things are when you're only $0.3$ of a child. Commented Feb 9, 2015 at 14:29
• @DavidK: Yes, I know -- the normal distribution is for continuous variables, not discrete ones. I'm not a statistician, but I assume there is some notion of a discrete random variable whose distribution is "normal-ish". If you can tell me how to express this using the official jargon, I'd be happy to reword the question. Commented Feb 9, 2015 at 23:22
• @bubba Technically I suppose a binomial distribution would do what you want--and for many purposes we would use a normal distribution as an approximation for it anyway. I don't think this detail (at least not by itself!) will stop anyone from giving a good answer. Commented Feb 10, 2015 at 0:26

Let $N$ be the number of machines and $X_n$ be the number of components of machine $n$. Then $$P(n\mbox{th Machine works sound})=\int_{-\infty}^\infty p^x \frac{1}{\sqrt{2\pi}\sigma}e^{(x-\mu)^2/{2\sigma^2}}dx=\int_{-\infty}^\infty e^{x\ln p} \frac{1}{\sqrt{2\pi}\sigma}e^{(x-\mu)^2/{2\sigma^2}}dx\\=p^\mu m_z(\sigma \ln p)=p^\mu e^{\sigma \mu \ln p+1/2\sigma^4(\ln p)^2}:=p'$$ where we have used the expression for the mgf of the normal distribution and $m_z(\cdot)$ is the mgf. Thus a machine picked at random will work sound with probability $p'$.
• Thanks. Your variables $N$ and $X_n$ are not used in your answer. What is $m_z$? Where does your second equality come from? Commented Feb 9, 2015 at 13:45
• Actually $N$ is used when I made the last statement, because there the probability becomes $\sum_{n=1}^NP$(the machine works sound|$n$th machine is chosen)P($n$th machine is chosen)=$\sum_{n=1}^N p'\cdot \frac{1}{N}=p'$. $m_z(\cdot)$ is the moment generating function for the standard normal distribution. Commented Feb 9, 2015 at 13:48