Answer
a. see explanations.
b. $0.99$
c. $0.01$
d. $0.41$
e. $23 $
Work Step by Step
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.
