Set theory and probability axioms

2 free worked examples

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1. Exam PSet theory and probability axiomsCore

   A motor insurer finds that 48% of its policyholders have a telematics device, 31% have a named second driver, and 17% have both. What proportion have at least one of the two?

   1. A$0.1700$
   2. B$0.6200$
   3. C$0.6500$
   4. D$0.7900$
   5. E$0.9600$

   Solution

   1. 'At least one' is the union, so inclusion-exclusion applies: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$.
   2. Substituting the given proportions gives $0.48+0.31-0.17$.
   3. That is $0.62$. The subtraction is essential because the 17% with both were counted once in each of the first two figures.
   4. As a check, the proportion with neither is $1-0.62=0.38$, which is consistent with the numbers given.

   Trap. Adding 0.48 and 0.31 to get 0.79 forgets that the overlap has been double-counted.

2. Exam PSet theory and probability axiomsExam level

   A risk register records $P(A)=0.5$, $P(B)=0.6$ and $P(A\cap B)=0.2$. Find the probability that exactly one of the two events occurs.

   1. A$0.1000$
   2. B$0.3000$
   3. C$0.5000$
   4. D$0.7000$
   5. E$0.9000$

   Solution

   1. 'Exactly one' is the union minus the intersection: the union counts everything that happens, and removing the overlap leaves the parts where only one occurred.
   2. $P(A\cup B)=0.5+0.6-0.2=0.9$.
   3. $P(\text{exactly one})=0.9-0.2=0.7$.
   4. Equivalently $\left[P(A)-P(A\cap B)\right]+\left[P(B)-P(A\cap B)\right]=0.3+0.4=0.7$, the same computation written as two disjoint pieces.

   Trap. Reporting the union (0.9), which includes the outcomes where BOTH events occur.