Question

4. Once upon a time there was King Jonathan of MARCsville. King Jonathan had a beautiful daughter whom he intended to marry to a prince from a neighboring kingdom. However, a short time before the proposed wedding day, the princess met Reynaldo, who was handsome, clever, romantic, but only a peasant. Their love flourished in secret. Inevitably King Jonathan heard, to his dismay, about their relationship. Irate, he ordered that Reynaldo be thrown into a room full of tigers. However, soft hearted King Jonathan offered a compromise: Reynaldo was to enter into an unfamiliar maze. At every intersection, he would have to choose which of the paths to follow. Eventually, he will reach one of two rooms. While the hungry tigers wait in one room, the loving princess waits in the other. King Jonathan showed a map of the maze to the princess and let her decide in which room to wait. Help the princess to decide by computing the probability that Reynaldo will end up in room A and the probability that Reynaldo will end up in room B. Assume that at each intersection, the chances for Reynaldo to choose any available paths are equal Es BI

Answer 1

Solution

Given that at each intersection, each available path has equal probability of being chosen,

P(E₁) = P(E₂) = P(E₃) = 1/3 ........................................................................................................................ (1)

P(E₄) = P(E₅) = ½ ...................................................................................................................................... (2)

P(E₆) = P(E₇) = P(E₈) = 1/3 ........................................................................................................................ (3)

Routes available to reach A and corresponding probabilities:

E₁ – E₄ : (1/3) x (1/2) = 1/6 [vide (1) and (2)]

E₃ – E₇ : (1/3) x (1/3) = 1/9 [vide (1) and (3)]

E₃ – E₈ : (1/3) x (1/3) = 1/9 [vide (1) and (3)]

Total probability = 1/6 + 2/9 = 7/18

Routes available to reach B and corresponding probabilities:

E₁ – E₅ : (1/3) x (1/2) = 1/6 [vide (1) and (2)]

E₃ – E₆ : (1/3) x (1/3) = 1/9 [vide (1) and (3)]

E₂ : 1/3 [vide (1)]

Total probability = 1/3 + 1/6 + 1/9 = 11/8

Thus, P(reaching A) = 7/18 and P(reaching B) = 11/8 Answer

[Going beyond,

Check: Reynoldo must reach either A and B and so P(A) + P(B) must be 1]

DONE