# Find the probability of a shaft going through a hole. Size distributions of shaft and hole are known

$r_1$ refers to the size of a shaft, and $r_2$ refers to a size of a hole, and their probability distributions are known. Failure is defined as when $r_1$ < $r_2$, i.e., when the shaft goes through the hole. The sizes of the shaft and that of the hole can be assumed to be normally distributed. I came across two ways, but not sure which one is correct:

1. Calculate the overlap area of the two probability distributions. However, I am not sure what the overlap area means in this case.
2. Calculate the failure probability, P($r_1$ < $r_2$) = $\int_0^\infty \int_0^{r_2} f(r_1,r_2)dr_1 dr_2$, where $f(r_1,r_2) = f_1(r_1)\: f_2(r_2)$ is the joint probability distribution of $r_1$ and $r_2$. (Formula taken from Ross S. M. Introduction to Probability and Statistics for Engineers and Scientists. Third Edition, Academic Press, 2004, p. 101.). I am not sure if the joint probability distribution means anything in this case, i.e., whether it is valid at all.

Also, is there any other way to calculate the failure probability?

• Please add the self-study tag and read its tag-wiki. Start by thinking of it in the simplest terms. If $Y$ is the r.v. representing the hole diameter and $X$ is the r.v. representing the shaft diameter, and assuming no need to allow room for "play" (which is unrealistic, but lets' go with it), then it fits when $X<Y$, or equivalently when $Y-X>0$. Consider first whether you should treat $X$ and $Y$ as independent (and if not, what's their joint distribution?). Then consider what the distribution of $Y-X$ must be. – Glen_b Aug 31 '17 at 1:33