Understanding the solution of a probability question 
Automobile policies are separated into two groups: low-risk and high-risk. Actuary Rahul examines low-risk policies, continuing until a policy with a claim is found and then stopping. Actuary Toby follows the same procedure with high-risk policies. Each low-risk policy has a 10% probability of having a claim. Each high-risk policy has a 20% probability of having a claim, The claim statuses of policies are mutually independent. Calculate the probability that Actuary Rahul examines fewer policies than Actuary Toby.

I found a solution online which begins as follows:
The probability that Actuary Rahul examines exactly $n$ policies $=0.9^{n-1}\times 0.1$.
The probability that Actuary Toby examines more than $n$ policies $=0.8^{n}$.
Then, the probability of both these events occurring is $=0.9^{n-1}\times 0.1 \times 0.8^{n}$ and the desired probability is given by $\sum_{n=1}^\infty 0.9^{n-1}\times 0.1 \times 0.8^{n}$.
My question is: in the second line of the solution, why is $0.8^{n}$ not the probability that Actuary Toby examines exactly $n+1$ policies? Is this online solution that I found even correct?

If Toby stops after examining exactly $n+1$ policies, this means that first $n$ policies would be non-claims, and the last (the $n+1$st policy) would be a claim. This has probability $0.8^n\times 0.2$. So $0.8^n$ is not the probability that Toby examines exactly $n+1$ policies.
The event that Toby examines more than $n$ policies is the same as the event that the first $n$ policies examined by Toby are all non-claims. This has probability $0.8^n$, as advertised.
