Chance the median of five random numbers tops 0.75

Question

Draw 5 independent numbers uniformly from [0,1]. The sample median is the 3rd smallest (the middle one). What is the probability that this median exceeds 0.75?

Accepted Answer

The median (3rd of 5) exceeds 0.75 exactly when at least 3 of the five numbers land above 0.75. Each number lands above 0.75 with probability 0.25, independently, so the count above 0.75 is Binomial(5, 0.25). We need 3, 4, or 5 of them: P = \binom{5}{3}\left(	frac14ight)^3\left(	frac34ight)^2 + \binom{5}{4}\left(	frac14ight)^4\left(	frac34ight) + \binom{5}{5}\left(	frac14ight)^5. Over the common denominator 4^5 = 1024, the three terms are \frac{90}{1024}, \frac{15}{1024}, and \frac{1}{1024}, so P = \frac{90 + 15 + 1}{1024} = \frac{106}{1024} = \frac{53}{512} \approx 0.104. Take…

Chance the median of five random numbers tops 0.75

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