Answer:
Step-by-step explanation:
for n ∈ N
Since Actuary Rahul examines low-risk policies, continuing until a policy with a claim is found and then stopping.
∴ the probability that Actuary Rahul examines exactly n policies
[tex](0.9)^{n-1}.(0.1)---(1)[/tex]
the probability that Actuary Toby examines more than  exactly n policies
[tex](0.8)^n---(2)[/tex]
Given that policies are actually independent
∴ the probability that the event  (1) and (2) happens simultaneously is
[tex](0.9)^{n-1}*(0.1)*(0.8)^n[/tex]
∴  the probability that Actuary Rahul examines fewer policies than Actuary Toby
[tex]\sum ^\infty _{n=1} (0.9)^{n-1}*(0.1)*(0.8)^n\\\\=(\frac{0.1}{0.9} \sum ^\infty _{n=1}(0.72)^n\\\\=\frac{1}{9} (\frac{0.72}{0.28} )\\\\=\frac{2}{7} \\\\=0.2857[/tex]