Problem 5. To control the amount of contamination particles in a semiconductor m
ID: 3044522 • Letter: P
Question
Problem 5. To control the amount of contamination particles in a semiconductor manufacturing facility, a process engineer conducted some experimental tests in lab environment of 450 cubic meters volume and found 1 contamination particle on average. Based on such information, evaluate the probability of observing zero contamination particle in an actual fabrication facility with a space of 2,250 cubic meters, and the probability of observing at most 2 contamination particles in this fabrication facilty,[20 points)Explanation / Answer
Given that, in 450 cubic meter there is 1 contamination particle on average. So, in 2250 cubic meter there would be 5 contamination particles on average (because 2250 = 450*5).
(1) Probability of no contamination particles = (1/2)5 = 1/32 = 0.03125
(2) Probability of atmost 2 contamination particles = probability of less than 2 particles
= P(0 particles) + P(1 particle)
= (1/32) + (5/32)
= 6/32 = 3/16 = 0.1875