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:
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$$p=0.6, \ q=0.4$$
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The team follows a binomial distribution $$B(18; 0,6)$$.
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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