probability

(a) Let the probability of final death by poison if the rat is currently in Cell 1, 2 and 3 be x, y and z respectivley.
The probability of death by poison when the rat is currently in cell 3 is:
1/5 chance goes directly to poison; and
1/5 chance goes to cell 1 and then finally death by poison; and
2/5 chance goes to cell 2 and then finally death by poison
Therefore z = 1/5 + x/5 + 2y/5
5z - x - 2y = 1 ... (1)
Similarly,
x = 1/4 + z/4 + 2y/4
4x - z - 2y = 1 ... (2)
y = 2x/5 + 2z/5
5y - 2x - 2z = 0 ... (3)
(1) - (2) => 6z - 5x = 0
x = 1.2z
Sub into(2), 4.8z - z - 2y = 1
y = 1.9z - 0.5
Sub into (3), 9.5z - 2.5 - 2.4z - 2z = 0
5.1z = 2.5
z = 25/51
The probability of death by poison if the rat is currently in cell 3 is 25/51
(b) f(x) = 2x; 0 <= x < = 1
Expected time to remain in a cell = ∫ xf(x) dx = 2/3
Let the expected time to death after entering cells 1,2 and 3 be a, b anc c respectively
c = the expected stay time in Cell 3 + (2/5) immediate death + (1/5) goes to cell 1 and accumulate the expected time to death there + (2/5) goes to cell 2 and accumulate the expected time to death there
c = 2/3 + (2/5)(0) + a/5 + 2b/5
15c - 3a - 6b = 10 ... (1)
Similarly,
a = 2/3 + (1/4)(0) + c/4 + 2b/4
12a - 3c - 6b = 8 ... (2)
b = 2/3 + (1/5)(0) + 2a/5 + 2c/5
15b - 6a - 6c = 10 ... (3)
(1) - (2) => 18c - 15a = 2
a = (18c - 2)/15
Sub into (1), 15c - (54c - 6)/15 - 6b = 10
225c - 54c + 6 - 90b = 150
b = (171c - 144)/90
b = (19c - 16)/10
Sub into (3), 15(19c - 16)/10 - 6(18c - 2)/15 - 6c = 10
28.5c - 24 - 7.2c + 0.8 - 6c = 10
15.3c = 33.2
c = 332/153 = 2.17 minutes