Memoryless property for Poisson process 
Suppose that arrivals of a certain Poisson process occur once every $4$ seconds on average. Given that there are no arrivals during the first $10$ seconds, what is the probability that there will be 6 arrivals during the subsequent $10$ seconds?

My solution
Suppose $N(t) =$ number of arrivals at time $t$. Then we need to find $$P(N(20) = 6 \mid N(10) = 0) = P(N(20) = 6) = e^{-\lambda t}\sum_{x=0}^6\frac{(\lambda t)^x}{x!} = e^{-80}\sum_{x=0}^6\frac{80^x}{x!}.$$
Since the fact that nothing occurs during the first $10$ seconds gives us no additional information.
 A: $$
P(N(20) = 6 \mid N(10) = 0) \ne P(N(20) = 6)
$$
$$
P(N(20) = 6 \mid N(10) = 0) = P(N(20) -N(10) = 6) = P(N(10) = 6) = \frac{e^{-\lambda\cdot 10} (\lambda\cdot10)^6}{6!}
$$
where $\lambda = \dfrac 1 {\text{4 seconds}} $.
It is not the Poisson distribution that is memoryless; it is the distribution of the waiting times in the Poisson process that is memoryless.  And that is an exponential distribution.
A: By the no-memory property of exponential inter-arrival times, you need to start over at the end of the
first 10 seconds. What did or did not happen in the first 10 seconds
is irrelevant to what happens in the second 10 seconds.
So your answer should be based on 10 sec. remaining.
Also, you need exactly 6 arrivals, not at least 6 arrivals.
The rate for a 10 second period of time is $\lambda = 10/4 = 2.5.$ 
If $X \sim \operatorname{Pois}(\lambda),$ then $P(X = 6) = e^{-\lambda} \lambda^6/6!.$
Related computations in R:
 lam = 10/4;  exp(-lam)*lam^6/factorial(6)
 ## 0.02783373
 dpois(6, lam)
 ## 0.02783373


