Three real numbers are chosen independently and uniformly at random from the interval [0, 3]. What is the probability that the smallest number is between 0.5 and 1?

Free worked solution · Quant Blueprint

61/216, approximately 28.24%.

1. Let M be the minimum of three independent uniform draws from [0, 3]. M > t exactly when all three draws exceed t.
2. Therefore P(M > t) = ((3 − t)/3)³ for t between zero and three.
3. Subtract the two tails: P(0.5 < M < 1) = (2.5/3)³ − (2/3)³ = 125/216 − 64/216 = 61/216.
4. The draws are continuous, so including or excluding the endpoints does not change the probability.

Common mistake

The minimum is not uniform. It is more likely to be near zero than near three.

Follow-up

For n independent draws from [0, L], use P(a < M < b) = (1 − a/L)ⁿ − (1 − b/L)ⁿ when 0 ≤ a ≤ b ≤ L.

Original educational explanation for this practice question. Reviewed 2026-10-13. Firm-style practice; the label describes the collection’s preparation context.