If errors in unedited manuscripts are Poisson distributed wi

If errors in unedited manuscripts are Poisson distributed with 3 errors in 5 pages and if each manuscript entails 4 pages, answer the following.

1: Probability of 1 or 3 errors in a manuscript

2: Probability of 4 or more errors in a manuscript

3: Probability of exactly 3.5 errors in a manuscript

4: Given three separate manuscripts, the probability that one of them will contain 4 or more errors and the other two will contain less than 4 errors

5: Given three combined manuscripts, the probability that the combination will contain more than 4 errors.

Solution

Possion Distribution
PMF of P.D is = f ( k ) = e- x / x!
Where   
= parameter of the distribution.
x = is the number of independent trials

3 errors in 5 pages
a)
error rate for manuscript entails 4 pages is 4*3/5, 2.4
P( X = 1 ) = e ^-2.4 * 2.4^1 / 1! = 0.2177
P( X = 3 ) = e ^-2.4 * 2.4^3 / 3! = 0.209

P( 1 or 3) = 0.2177+0.209 = 0.4267
b)

P( X < 4) = P(X=3) + P(X=2) + P(X=1) + P(X=0)   
= e^-2.4 * 3 ^ 3 / 3! + e^-2.4 * ^ 2 / 2! + e^-2.4 * ^ 1 / 1! + e^-2.4 * ^ 0 / 0!
= 0.7787
P( X > = 4 ) = 1 - P (X < 4) = 0.2213

c)
P( X = 3.5 ) = e ^-2.4 * 2.4^3.5 / 3.5! = 0.209