Probability of 3 people winning 3 games out of 9 total. 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 $3$}) &amp; = \binom{9}{3}(0.4^3)(0.6^6)\\
P(\text{Bill wins $3$}) &amp; = \binom{9}{3}(0.3^3)(0.6^6)\\
P(\text{Cara wins $3$}) &amp; = \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.
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
$$\frac{9!}{3!3!3!0!}.4^3.3^3.2^3.1^0=0.02322432=2.32%$$
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
$$\frac{9!}{3!3!3!0!}.4^3.3^3.2^3.1^0=0.02322432=2.32%$$