Question

A machine prints a word and the number of letters in this word is a Poisson distributed random variable with parameter λ (so it could possibly have zero letters). However, each letter in the word is printed incorrectly with probability 2/3 independently of all other letters. Compute the expectation and the variance of the number of incorrect letters in the word that the machine prints.

Answer 1

Given:

A machine prints a word and the number of letters in this word is a Poisson distributed random variable with parameter λ .

However, each letter in the word is printed incorrectly with probability 2/3 independently of all other letters.

So probability of world is printed incorrectly is

p(Incorrect) = 2/3 = 0.667

Mean, =

Standard deviation, = √

So variance of Poisson =

Variance of probability = p(1-p) = 2/3(1-2/3) = 2/9

1) The expectation of the number of incorrect letters in the word that the machine prints is

expectaion = Mean × P(Incorrect)

= λ×2/3

= 2λ/3

2) The variance of the number of incorrect letters in the word that the machine prints.

Variance = variance of poisson * variance of probability

= λ × p(1-p)

= λ × 2/9

= 2λ/9