A basketball team is in a very close league with $$10$$ teams. We can consider that the probability of winning is the same in each of the $$18$$ games.
 To define a victory probability ($$p$$) and one of defeat ($$q$$), so that the results are quite balanced.
 What distribution correctly models the behavior of the team?
 What is the probability that the team wins exactly ten games? And the probability of winning all of them? And none of them?
 What is the average number of victories per season if the team plays several years under these circumstances?
See development and solution
Development:

$$p=0.6, \ q=0.4$$

The team follows a binomial distribution $$B(18; 0,6)$$.

Using the probability function of a binomial distribution: $$$p(X=10)=\binom{18}{10}\cdot0,6^{10}\cdot0,4^8$$$ $$$p(X=10)=\dfrac{18!}{10!\cdot8!}\cdot0,6^{10}\cdot0,4^8=0,173$$$ $$$p(X=0)=\binom{18}{0}\cdot0,6^{0}\cdot0,4^{18}$$$ $$$p(X=0)=1\cdot0,6^0\cdot0,4^{18}=6,87\cdot10^{8}$$$ $$$p(X=18)=\binom{18}{18}\cdot0,6^{18}\cdot0,4^0$$$ $$$p(X=18)=\dfrac{18!}{18!}\cdot0,6^{18}\cdot1=1,01\cdot10^{4}$$$
 $$\mu=18\cdot0,6=10,8$$ wins
Solution:
 $$p=0.6, \ q=0.4$$
 The team follows a binomial distribution $$B(18; 0,6)$$.
 $$p(X=10)=0,173$$; $$p(X=0)=6,87\cdot10^{8}$$; $$p(X=18)=1,01\cdot10^{4}$$
 $$\mu=18\cdot0,6=10,8$$ wins