The binomial (or Bernoulli) distribution

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.

  1. To define a victory probability ($p$) and one of defeat ($q$), so that the results are quite balanced.
  2. What distribution correctly models the behavior of the team?
  3. What is the probability that the team wins exactly ten games? And the probability of winning all of them? And none of them?
  4. What is the average number of victories per season if the team plays several years under these circumstances?

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

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

  3. Using the probability function of a binomial distribution: $$p(X=10)=\binom{18}{10}\cdot0,6^{10}\cdot0,4^8$$ $$p(X=10)=\dfrac{18!}{10!\cdot8!}\cdot0,6^{10}\cdot0,4^8=0,173$$ $$p(X=0)=\binom{18}{0}\cdot0,6^{0}\cdot0,4^{18}$$ $$p(X=0)=1\cdot0,6^0\cdot0,4^{18}=6,87\cdot10^{-8}$$ $$p(X=18)=\binom{18}{18}\cdot0,6^{18}\cdot0,4^0$$ $$p(X=18)=\dfrac{18!}{18!}\cdot0,6^{18}\cdot1=1,01\cdot10^{-4}$$

  4. $\mu=18\cdot0,6=10,8$ wins

  1. $p=0.6, \ q=0.4$
  2. The team follows a binomial distribution $B(18; 0,6)$.
  3. $p(X=10)=0,173$; $p(X=0)=6,87\cdot10^{-8}$; $p(X=18)=1,01\cdot10^{-4}$
  4. $\mu=18\cdot0,6=10,8$ wins
Back to topic