## Precalculus (10th Edition)

a) $0.45$ b)$0.75$ c)$0.9$
These events are mutually exclusive hence I can obtain the required results by simply adding the individual probabilities. We know that probability$=\frac{\text{number of favourable outcomes}}{\text{number of all outcomes}}.$ a) P(at most 2 people)=$P(0)+P(1)+P(2)=0.10+0.15+0.20=0.45$ b) P(at least 2 people)=$P(2)+P(3)+P(\geq4)=0.20+0.24+0.31=0.75$ c)P(at least 1 person)=$P(1)+P(2)+P(3)+P(\geq4)=0.15+0.20+0.24+0.31=0.9$