Customers arrive at a shop according to a Poisson process at rate $\lambda$ (/minute), where they choose to buy either product $A$ (with probability p) or product $B$ (with probability $1-p$), independently. Given that during the first hour $5$ customers chose product $B$, what is the probability all the customers that arrived at the shop within the first $10$ minutes, all bought product $A$?
Attempt. Let $N_A(t), N_B(t)$ the the Poisson processes, of rates $\lambda p, ~\lambda(1-p)$, respectively, that count the produts of type $A$ and $B$ bought, respectively, at time $t$. Then the probability we are asked to find is:
$$\sum_{k=0}^{\infty}P(N_A(10)=k~|~N_A(10)+N_B(10)=k)\cdot P(N_A(10)+N_B(10)=k~|~N_A(60)+N_B(60)=k+5)\cdot P(N_A(60)+N_B(60)=k+5~|~N_B(60)=5),$$
where $$P(N_A(10)=k~|~N_A(10)+N_B(10)=k)=\binom{k}{k}\Big(\frac{\lambda p}{\lambda p+\lambda(1-p)}\Big)^k\Big(1-\frac{\lambda p}{\lambda p+\lambda(1-p)}\Big)^{k-k}=p^k,$$
$$P(N_A(10)+N_B(10)=k~|~N_A(60)+N_B(60)=k+5)= \binom{k+5}{k}\Big(\frac{10}{60}\Big)^k\Big(1-\frac{10}{60}\Big)^{5}$$
and
$$P(N_A(60)+N_B(60)=k+5~|~N_B(60)=5)= P(N_A(60)=k~|~N_B(10)=5)=P(N_A(60)=k)= e^{-60\lambda p}\frac{(60\lambda p)^k}{k!}.$$
Am I on the right path? Thanks a lot in advance!
The direct approach here would be to note that all customers who arrived in the first $10$ minutes buying $A$ is the same as none of them buying $B$, so we don't have to worry about the customers who bought $A$ or about the whole Poisson business at all, since we just want the probability that of $5$ customers equidistributed over $60$ minutes none arrived in the first $10$ minutes, which is
$$ \left(\frac{50}{60}\right)^5=\left(\frac56\right)^5=\frac{3125}{7776}\approx40\%\;. $$