X is said to be Position distributed with parameter lambda >0 if it has the following pmf: PX(x)= (lambda ^x)*e^(-lambda)/ x! where x = 0, 1, 2... and 0 otherwise.
1. For a certain virus , it is discovered that in 74% of the cases where the virus replicates, the replicate contains at least one mutation. If we assume that the number of mutations per replication is a random variable having a Poisson distribution, find the mean number of mutations per replication, and the percentage of replications that have 5 or more mutations.
2. Suppose that X~ Poisson(lambda), find E[X(X-1)(X-2)(X-3)].