Find Probability of hitting 1 gold grain in a 1m cube with 260 randomly located grains of gold.?

Question

There are 260 randomly distributed gold grains each of 1 cu.mm size in a 1 cu.m. cube of rock. If I drill a cylinder of 30mm radius through an axis of the block, normal to a face, what is the probalility of getting 
1. No gold grain?
2. One gold grain? 
3. Two gold grains?
A real life problem!!

nealjking · Accepted Answer

If you drill a cylinder of radius 30 mm = 3e-2 meters through a side, you get a volume of pi*(3e-2)^2 * 1 = 9*pi*e-4. 

This number, divided by the volume of the cubic block, is the probability of getting any specific one grain. Let's call it p:
p = 9*pi*e-4/1 = 9*pi*e-4 . 
p = 2.827e-3

1) What is the probability of getting 0 grains? 
That would mean that you had 260 shots at getting a grain, and lost every time. Since the probability of missing a particular grain is 1-p, the probability of missing all 260 is 
(1-p)^260 = (0.9972)^260 = 0.4789

2) What is the probability of getting 1 grain?
That would mean missing 259 times, but hitting exactly once. Since you get 260 choices to hit, the probability is:
260*p*(1-p)^259 = 260*(2.827e-3)*(0.9972)^259
                              = 0.3531

3) What is the probability of getting 2 grains?
What we are clearly getting here is the binomial distribution, so the answer is
C(260; 2) * p^2 * (1-p)^258
= ((260)(259)/2)(2.827e-3)^2 * (0.9972)^258
= 0.1297

Anonymous · Answer

What is the average size/volume of a gold grain?

Find Probability of hitting 1 gold grain in a 1m cube with 260 randomly located grains of gold.?

2 Answers