Median of three random numbers above 0.6

Question

Draw 3 independent numbers uniformly from [0,1]. The sample median is the 2nd smallest (the middle one). What is the probability that this median exceeds 0.6?

Accepted Answer

The median (2nd of 3) exceeds 0.6 exactly when at least 2 of the three numbers land above 0.6. Each lands above 0.6 with probability 0.4, so the count above 0.6 is Binomial(3, 0.4). We need 2 or 3: P = \binom{3}{2}(0.4)^2(0.6) + \binom{3}{3}(0.4)^3 = \frac{36}{125} + \frac{8}{125} = \frac{44}{125} = 0.352. Takeaway. For an odd sample of three, the median tops t iff at least two of the three exceed t: a quick binomial with n=3.

Median of three random numbers above 0.6

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