The answer is 5, geometric distribution with p=0.2.
Recognize this as a geometric distribution problem where each draw is a Bernoulli trial with success probability p = 0.2 (since P(X > 0.8) = 0.2). The expected number of trials until the first success is 1/p = 5. Explain the reasoning clearly, including the memorylessness property and the formula for the mean of a geometric distribution.
Pro tip: Connect this to product analytics by mentioning how such calculations inform experiment design, e.g., estimating the number of user actions needed to observe a rare event, and note that the result is independent of the specific threshold as long as the probability is correctly computed.
Define a 'success' as drawing a value greater than 0.8. Since each draw is independent and has constant success probability, the number of draws until the first success follows a geometric distribution.
For X ~ Uniform(0,1), P(X > 0.8) = 1 - 0.8 = 0.2. So p = 0.2.
The expected number of trials for the first success in a geometric distribution is E[N] = 1/p. Thus, E[N] = 1/0.2 = 5.
Interpret the result: on average, it takes 5 draws to get a value > 0.8. Validate by considering that each draw has a 20% chance, so intuitively about 1 in 5 draws should succeed.
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