Poisson Random Variable

Quote:

Originally Posted by essedra

The number of defective items that come out of a production line on any given day is a Poisson random variable with parameter λ=2. At the end of the day, the defective items are reworked. Each defective item can be repaired with probability 0.6 and is discarded with probability 0.4.

What is the probability that fewer than 3 items are discarded on a given day?
What is the expected number of items discarded?

The probability of $k$ defectives in a day is:

$<br /> p(\text{k})=f(k,\lambda)=f(k,2)<br />$

where $f(k,\lambda)$ is the Poisson probability mass function.

Hence the probability that fewer that 3 items are discarded in a day is:

$p(\text{fewer than 3 discards})=\sum_{k=0}^{\infty} f(k,2)\sum_{r=0}^2 b(r;k,0.4)$

where $b(r;k,0.4)$ is the pmf for the binomial distribution with k trials with probability of success on a single trial of $0.4$ .

CB