2 . This fails to occur if and only if the disjoint union A0 ∪ A1 ∪ · · · ∪ An occurs, where A0 is the event there is no break in (0, 1 2 ], and Ak is the event of no break in (Xk , Xk + 1 2 ] for k ≥ 1 (remember the permanent break at 1). Now, P(A0) = ( 1 2 )n, and for k ≥ 1, P(Ak) =

Z 1 0

P(Ak | Xk = x) dx =

Z 1 2

0

( 1 2 )n−1 dx = ( 1 2 )n.