Applied probability

In fact, after 5 seconds, both X and Y can only be in the odd units from the origin.

That is, X and Y can only be at -5, -3, -1, 1, 3 or 5 units from the origin. Take to the right be positive.

To calculate the probability that Y will be in a position to the right of X, we can skip calculating the probabilities that X is in +5 units, and Y is in -5 units.

P(X is at +3 units) = P(X walks 4 times right and 1 time left)

= 5C1(2/3)4(1/3) = 80/243

P(X is at +1 unit) = P(X walks 3 times right and 2 times left) = 5C2(2/3)3(1/3)2 = 80/243

P(X is at -1 unit) = P(X walks 2 times right and 3 times left)
= 5C3(2/3)2(1/3)3 = 40/243

P(X is at -3 units) = P(X walks 1 time right and 4 times left)
= 5C4(2/3)(1/3)4 = 10/243

P(X is at -5 units) = P(X walks all to the left) = (1/3)5 = 1/243

P(Y is at +5 units) = P(Y walks all to the right) = (1/2)5 = 1/32

P(Y is at +3 units) = P(Y walks 4 times right and 1 time left) = 5C1(1/2)4(1/2) = 5/32

P(Y is at +1 unit) = P(Y walks 3 times right and 2 times left)
= 5C2(1/2)3(1/2)2 = 5/16

P(Y is at -1 unit) = P(Y walks 2 times right and 3 times left)
= 5C3(1/2)2(1/2)3 = 5/16

P(Y is at -3 units) = P(Y walks 1 time right and 4 times left)
= 5C4(1/2)(1/2)4 = 5/32

Therefore, for Y to be right of X,

Case I: X is at -5 units, then the probability = (1/243)(1/32 + 5/32 + 5/16 + 5/16 + 5/32) = 31/7776

Case II: X is at -3 units, then the probability = (10/243)(1/32 + 5/32 + 5/16 + 5/16) = 65/1944

Case III: X is at -1 unit, then the probability = (40/243)(1/32 + 5/32 + 5/16) = 20/243

Case IV: X is at +1 unit, then the probability = (80/243)(1/32 + 5/32) = 5/81

Case V: X is at +3 units, then the probability = (80/243)(1/32) = 5/486

So, the required probability is summing up the probabilities of all possible cases, it is 497 / 2592.