6 Suppose that a certain type of magnetic tape contains on t

6. Suppose that a certain type of magnetic tape contains on the average three defects per 1000 feet. What is the probability that a roll of tape 1200 feet long contains no defects?

Solution

Let X denote the number of defects in one foot of tape.
Then X ? Poisson(?),
where ? = 3 1000 = 0.003.
It is reasonable to assume that the number of defects in a foot of the tape is independent of the number of defects in any other foot of the tape.
Thus, if we let Y denote the number of defects in 1200 feet of the tape,
Y ? Poisson(1200?) or Poisson(3.6).
Thus,
Pr(Y = k) = (3.6)^k / k! * e^-3.6 for k = 0, 1, 2, 3, . . . ;
Pr(Y = k) = zero elsewhere.
Consequently, the probability of no defects in the 1200 feet of tape is
Pr(Y = 0) = e^-3.6 or about 2.73%.