Question

Consider a line axis where there are blocks at position x = 3 and x = -3 and a point moves along the axis. Every time the point at position x (-3 < x < 3) can either move left (x - 1) with probability of 0.5 or move right (x + 1) with probability of 0.5. However, it cannot penetrate the block (e.g., at position x = 3, if the point keeps moving right, it stays in the same position. This situation still counts one move). Starting from x = 0, what is the probability of the point being at position x = 0 after 8 moves?

Answer 1

$Let~X~be~a~random~variable~denoting~no.~of~right~moves~after~8~moves.\\ Then~X\sim Binomial(8,0.5).\\ Required~probability=P(X=4)=\binom{8}{4}(0.5)^8=0.2734$