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