Four quick waits in a row

Question

You make four independent visits to a kiosk. Each visit's wait time is equally likely to be anywhere between 0 and 10 minutes (a uniform draw), and the visits do not affect each other.

What is the probability that all four waits are under 5 minutes?

Accepted Answer

Take one visit first. For a single uniform wait on 0 to 10 minutes, the chance it is under 5 is the fraction of the range below 5: P(	ext{one wait} < 5) = \frac{5}{10} = \frac12. Since the four visits are independent, multiply the individual chances to get the chance they all stay short: P(	ext{all four} < 5) = \left(\frac12ight)^4 = \frac{1}{16} = 0.0625. This is exactly the probability that the longest of the four waits is under 5 minutes, because "all four short" and "the maximum is short" are the same event. Each extra visit halves the odds again, which is why going from three draws (…

Four quick waits in a row

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