Question

Two chess players have the probability Player A would win is 0.40, Player B would win is 0.35, game would end in a draw is 0.25. If these two chess players played 12 games, what is the probability that Player A would win 7 games, Player B would win 2 games, the remaining 3 games would be drawn?

Answer 1

Multinomial Distribution

$$p=\dfrac{n!}{(n_1)!(n_2)!(n_3)!}(p_1)^{n_1}(p_2)^{n_2}(p_3)^{n_3}$$

Given

$n=12$

$n_1=7$

$n_2=2$

$n_3=3$

$p_1=0.40$

$p_2=0.35$

$p_3=0.25$

$$p=\dfrac{12!}{(7)!(2)!(3)!}(0.40)^{7}(0.35)^{2}(0.25)^{3}$$ $$=0.02483712$$