A sieve is made of a square mesh of wires. Each wire has diameter d, and the holes in the mesh are squares whose side length is w. A spherical particle of radius r is dropped on the mesh. What is the probability that it fails to pass through if it is dropped n times? (Calculations such as these are relevant to the theory of sieving for analyzing the size distribution of particulate matter.)
my answer is $\frac{(w-d-2r)^2}{(w-d)^2}$ am i right?help,i dont't known how to proceed
If $w$ is the width of a mesh hole, then $(\frac{1}{2}d+w+\frac{1}{2}d)=(w+d)$ is the distance from one wire midpoint to the midpoint of the wire on the opposite side of the hole. If $N$ is the number of holes, then the total area of the mesh is
$$A_{total} = N\, (w+d)^2$$
In order to pass through the hole without touching a wire, the center of the spherical particle needs to be positioned within a smaller square that has the side $(w-2r)$. The total area that allows pass-through is therefore
$$A_{pass} = N\, (w-2r)^2$$
For a single drop, the probability of passing through is
$$P_{pass} = \frac{A_{pass}}{A_{total}} = \frac{(w-2r)^2}{(w+d)^2}$$
And the probability of failing is therefore
$$P_{fail} = 1 - P_{pass}$$
Assuming each drop is an independent event, the probability of failing $n$ drops is
$$(P_{fail})^n = (1-P_{pass})^n = \left(1-\dfrac{(w-2r)^2}{(w+d)^2}\right)^n$$