## Precalculus (6th Edition) Blitzer

a. see explanations. b. $0.99$ c. $0.01$ d. $0.41$ e. $23$
a. Imagine the process of choosing two birthdays. The first choice can be any day; that is, $p_1=1=\frac{365}{365}$. The second choice can be any day but the day of the first choice; that is, $p_2=\frac{364}{365}$. Thus, the probability that they do not have the same birthday is $\frac{365}{365}\times\frac{364}{365}$. b. Using the similar argument in part-(a), we have $p=\frac{365}{365}\times\frac{364}{365}\times\frac{363}{365}\approx0.99$ c. This should be the complement of part-(b); thus $p'=1-0.99=0.01$ d. First find the probability that all 20 people have different birthdays $p=\frac{365}{365}\times\frac{364}{365}\times\frac{363}{365}\times\frac{346}{365}\approx0.59$ Then, the complement is $p'=1-p=0.41$ e. We need $p'\geq 0.5$ or $p\leq 0.5$. Start from 20, increase the number of people and test until 23. We have $p=\frac{365}{365}\times\frac{364}{365}\times\frac{363}{365}\times\frac{343}{365}\approx0.49$ and $p'=1-p=0.51$ That is, we need 23 or more people.