One cut, a dominant piece

Question

A tape of length 1 is cut at a single uniformly random point, producing two pieces. What is the probability that the longer piece makes up more than 70\% of the whole tape?

Accepted Answer

Let the cut be at U \sim U(0, 1); the longer piece is \max(U, 1 - U). The longer piece exceeds 0.7 exactly when the cut is far enough toward one end: U 0.7 \quad	ext{or}\quad U < 0.3 . Each interval has length 0.3, so P = 0.3 + 0.3 = 0.6 = 	frac35 . Takeaway. "Longer piece above fraction f" (with f 	frac12) means the cut lands in the outer 1 - f at each end, probability 2(1 - f). Here 2(0.3) = 	frac35.

One cut, a dominant piece

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