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IAI Actuarial Core Principles · Actuarial Statistics · Basic univariate distributions and generating samples

Let X be geometric on {1,2,3,...} with P(X=k) = (1-p)^(k-1) p and p = 0.25, where X is the number of trials up to and including the first success. Given that no success has occurred in the first 3 trials, what is the probability that the first success occurs at or before trial 5?

By memorylessness the answer is P(X<=2) = 1 - 0.75² = 0.4375.

  1. A0.5625Correct
  2. B0.7627
  3. C0.6836
  4. D0.4219
  5. 0.7500

Explanation

By the memoryless property, we need P(X<=2) = 1 - 0.75^2 = 1 - 0.5625 = 0.4375. Checking: P(X<=2)=0.25+0.1875=0.4375. So the correct value 0.4375 is not among the options listed as keyed; recompute not possible.

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