Discrete distributions

2 free worked examples

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1. Exam PDiscrete distributionsCore

   A claims handler reviews 10 independent files, each incorrectly coded with probability 0.25. What is the probability exactly 3 are incorrectly coded?

   1. A$0.0156$
   2. B$0.1877$
   3. C$0.2503$
   4. D$0.2816$
   5. E$0.7500$

   Solution

   1. Fixed number of independent trials with a constant success probability makes this binomial with $n=10$, $p=0.25$.
   2. $P(X=3) = \binom{10}{3}(0.25)^{3}(0.75)^{7}$.
   3. $= 120 \times 0.015625 \times 0.133484$.
   4. $= 0.2503$.
2. Exam PDiscrete distributionsExam level

   The number of claims filed by a policyholder takes values 0, 1, 2, 3 with probabilities 0.25, 0.35, 0.28 and 0.12. Find the mean.

   1. A$0.9371$
   2. B$1.0000$
   3. C$1.2700$
   4. D$1.5000$
   5. E$2.5500$

   Solution

   1. $E[X] = \sum x\,p(x)$ over the support.
   2. $= 0(0.25) + 1(0.35) + 2(0.28) + 3(0.12)$.
   3. $= 0 + 0.35 + 0.56 + 0.36$.
   4. $= 1.27$ claims per policyholder per year.
   5. The mean is not one of the attainable values, which is normal: an expectation is a weighted centre of mass, not an outcome.
   6. It sits above 1 because the upper values, though less likely, pull the average up.