Question

Solve the problem. Arat is placed at random in one of five compartments in a maze. Compartments 1, 2, and 5 have food; so the rat will remain at any of these once reached. If the rat is placed in compartment 3, what is the expected number of times it will be in compartment 3 or 4 before it finds food? Round your answer to the nearest hundredth. 1 12345 1 0 0 0 2 0 1 0 0 3 0.22 0.13 0.16 0.25 4 0.22 0.17 0.14 0.15 5 0 0 0 0 0 0 0.24 0.32 1 13 O 1.62 22 O 16

Answer 1

Let a and b be the expected number of times it will be in compartments 3 or 4 uptill finding food, given it started from 3 and 4 respectively. Thus, a = 0.16(a) +0.25(b)+(1 – 0.16 – 0.25) x 1+1 = 0.16a+0.256 +1.59 Similarly, b = 0.14(a) +0.15(6) + (1 – 0.14 – 0.15) x 1+1 = 0.14a +0.156 + 1.71 The above two equations are: 0.84a – 0.25b = 1.59 – 0.14a + 0.85b = 1.71 Eliminating b, we get: (0.84 x 0.85 – 0.14 x 0.25)a = (1.59 x 0.85 +1.71 x 0.25) 0.679a 1.779 a= 2.62 Now in expected number of steps up till finding food is 2.62. Thus, the expected number of steps before finding food is: 2.62 – 1= 1.62 (Ans.)