N/10= 1-N/10 at N =5; so when the frog is aat lily pad 5 it has equal probability of either being eaten up by the snake or get escaped. Let (P_k) represents the probability that the frog escapes if it is currently on pad k then (P_5)= 1/2.
Solving the following equations
(P_1)= 9/10 (P_2)
(P_2)= 2/10 (P_1)+8/10 (P_3)
(P_3)= 3/10(P_2)+ 7/10 (P_4)
(P_4)= 4/10 (P_3)+6/10 (P_5)
(P_5)= 1/2
SOLVING THE ABOVE EQUATIONS STEP BY STEP WE GET
\(P_4\)= 2/5 \(P_3\)+ 3/10
(P_3)= 3/10 (P_2) +7/10 ×(2/5 (P_3 +3/10))
(P_3)= 3/10 (P_2) + 14/50 (P_3) +21/100
36/50 (P_3)= 3/10 (P_2) + 21/100
\P_3)= (3/10 (P_2) +21/100)×50/36
(P_2)= 2/10 (P_1) + 8/10 ( 3/10 (P_2) +21/100)×50/36
= 2/10 (P_1) +1/3 (P_2) + 7/30
2/3 (P_2)=2/10 (P_1)+7/30
(P_2) = 3/10 (P_1) + 7/20
(P_1) = 9/10×(3/10 (P_1) + 7/20)
(P_1)= 27/100 (P_1) +63/200
(P_1)×73/100= 63/200
(P_1)= 63/146
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