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I have a question and I'm not sure if my answer is correct. Suppose Al, Bill, Cara, and Bob play cards together. Al wins with probability $0.4$, Bill with $0.3$, Cara with $0.2$, and Bob with $0.1$. Assume they play 9 games total and the outcomes of each game are independent. What is the probability Al, Bill, and Cara win $3$ games each and Bob does not win any? My answer: \begin{align*} P(\text{Al wins $3$}) & = \binom{9}{3}(0.4^3)(0.6^6)\\ P(\text{Bill wins $3$}) & = \binom{9}{3}(0.3^3)(0.6^6)\\ P(\text{Cara wins $3$}) & = \binom{9}{3}(0.2^3)(0.8^6) \end{align*} I know they're independent events so I added them: $$P(\text{Al wins $3$}) + P(\text{Bill wins $3$}) + P(\text{Cara wins $3$})$$

I don't have a solution, and I'm not sure if this is correct reasoning.

EDIT: The correct answer is to use the multinomial probability.

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    $\begingroup$ "I know they're independent events so I added them" There is no situation in which this would be a correct choice to do for independent events. Do not confuse "independent" with "mutually exclusive". Further, the event that Al wins 3 is not independent of the event that Bill wins 3. What is independent here is the event that Al wins the first game and Bill wins the second game and other similar pairs of events. $\endgroup$ – JMoravitz Oct 24 '17 at 19:49
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    $\begingroup$ You did calculate the proability that Al wins 3 correctly, but that isn't particularly useful for our problem. As for a correct approach, consider the multinomial distribution. $\endgroup$ – JMoravitz Oct 24 '17 at 19:50
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    $\begingroup$ @JMoravitz Thank you! I see now that I want to use the multinomial probability with trials N = 9; four categories of n1 = 3, n2 = 3, n3 = 3, n4 =0; with P1 = .4, P2 = .3, P3 = .2, P4 = .1; and plug in getting P = [ 9!/(3!)(3!)(3!)(0!) ] * [ (.4^3)(.3^3)(.2^3)(.1^0) ] $\endgroup$ – Tom Oct 24 '17 at 20:05
  • $\begingroup$ Welcome to MathSE. Here is a tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Oct 24 '17 at 20:42
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Using multinomial probability, the formula is the following: $$p=\frac{n!}{n_1!n_2!n_3!\dots}p_1^{n_1}p_2^{n_2}p_3^{n_3}p_4^{n_4}\dots$$ You use this because each of the events (or games) are independent of each other and the probability of one doesn't affect the other. You also cannot use binomial because there are more than 2 factors (3 people playing).

$n$ is the number of trials. $n_i$ are the individual events, the number of times Al, Bill, Cara, and Bob win the game. Al, Bill, and Cara's events are 3, since they need to win 3 times. Bobs event is 0 since he is not winning. Then we have the answer as $$\frac{9!}{3!3!3!0!}.4^3.3^3.2^3.1^0=0.02322432=2.32%$$

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