Poisson Distribution Question and Memoryless property I have this question assigned to me
Suppose that car accidents in a given town have a Poisson distribution, with an average of 5 per week (i.e. λ = 5). Also, note that, given a Poisson process with mean λ, the waiting time until the first outcome has an exponential distribution with mean  1/λ .
Suppose that no accidents have occurred for two weeks. What is the probability that no accidents will occur for another two weeks? Give the exact value or round your answer to six decimal places.
When I first read this, I interpreted it as asking what is the probability that no accidents will occur for 4 weeks, given there were no accidents in the first two weeks.
I executed the following simple code in R using the dpois function to come up with my answer
#k = 0 because that is the provided number of "successes" we would like to determine"
#lamda = 20 because the rate of accidents is 5 a week over 4 weeks would be 20
dpois(0,20,log = FALSE)

However this does not give the right answer. The correct answer is given by executing
dpois(0,10,log = FALSE)

I believe this has something to do with the memoryless property which is a little lost on me. Is there someone that can help explain what I am missing here?
 A: 
what is the probability that no accidents will occur for 4 weeks, given there were no accidents in the first two weeks.

Correct.

dpois(0,20,log = FALSE) does not give the right answer

Indeed, because this is not the conditional probability you intended to calculate. This is just the unconditional probability of no accidents over four weeks.
To obtain the conditional probability you need to divide by the probability of the conditioning event, which is ... ?


The correct answer is given by executing
dpois(0,10,log = FALSE)
I believe this has something to do with the memoryless property which is a little lost on me. Is there someone that can help explain what I am missing here?

Indeed you are right, but from the point of view of Poisson processes we don't always call it the memoryless property, it's rather part of the definition:
$P(\text{no accidents in weeks 3 and 4 }|\text{ no accidents in weeks 1 and 2}) = P(\text{no accidents in weeks 3 and 4})$ because the counts of the Poisson process over two disjoint periods of time are independent.
From the point of view of exponential distributions, this is the memoryless property, $P(X>\text{4 weeks} | X\text{> 2 weeks}) = P(X \text{> 2 weeks})$
A: The question is asking "Suppose that no accidents have occurred for two weeks. What is the probability that no accidents will occur for another two weeks?". The sentence starting "Suppose that ..." is giving us prior information, in other words we know that there have been no accidents in the previous two weeks. In probability terms, that means we can condition on the event "no accidents have occurred for two weeks".
Then we're asking for the probability that no accidents occur in the next two weeks. So you're right in interpreting the question as asking for "the probability that no accidents will occur for 4 weeks, given there were no accidents in the first two weeks", but the problem was that you just calculated the probability of no accidents in 4 weeks without applying the given condition.
If we say that $X_1, X_2, X_3, X_4$ are the number of accidents happening in the last two and next two weeks, then we're looking for $P(X_1 + X_2 + X_3 + X_4  = 0 | X_1 + X_2 = 0)$. Because the Poisson processes are memoryless, though, $X_3$ and $X_4$ are independent of $X_1$ and $X_2$ and so that just becomes $P(X_3 + X_4 = 0)$ to which you can then apply $X_3 + X_4 \sim Poisson(2\lambda)$ to get the given solution.
