Using the formula for the multinomial distribution, the probability of $X _ { 1 } = 2 , X _ { 2 } = 1 , X _ { 3 } = 2$ when $\mathrm { n } = 5$, $\mathrm { p } _ { 1 } = 0.4 , \mathrm { p } _ { 2 } = 0.5 , \mathrm { p } _ { 3 } = 0.1$ is
A) $\{ 5 ! / ( 2 ! \times 1 ! \times 2 ! ) \} _ { 2 } \times 0.4 ^ { 2 } \times 0.5 ^ { 1 } \times 0.1 ^ { 2 }$
B) $2 ! \times 1 ! \times 2 ! \times 0.4 ^ { 2 } \times 0.5 ^ { 1 } \times 0.1 ^ { 2 }$
C) $5 ! \times 0.4 ^ { 2 } \times 0.5 ^ { 1 } \times 0.1 ^ { 2 }$
D) $\{ 5 ! / ( 0.4 ! \times 0.5 ! \times 0.1 ! ) \} \times 2 ^ { 0.4 } \times 1 ^ { 0.5 } \times 2 ^ { 0.1 }$

Question

Quizplus · Accepted Answer

The formula for the multinomial distribution is:
$$P(X_1=x_1,X_2=x_2,...,X_k=x_k) = \frac{n!}{x_1!x_2!...x_k!}p_1^{x_1}p_2^{x_2}...p_k^{x_k}$$
where $n$ is the total number of trials, $x_i$ is the number of successes in the $i$th category, $p_i$ is the probability of success in the $i$th category, and $k$ is the total number of categories.

In this problem, we have $k=3$, $n=5$, $x_1=2$, $x_2=1$, $x_3=2$, $p_1=0.4$, $p_2=0.5$, and $p_3=0.1$. Plugging these values into the formula, we get:
$$P(X_1=2,X_2=1,X_3=2) = \frac{5!}{2!1!2!}0.4^2 0.5^1 0.1^2$$
Simplifying, we get:
$$P(X_1=2,X_2=1,X_3=2) = \{ 5! / ( 2 ! 	imes 1 ! 	imes 2 ! ) \}_{2} 	imes 0.4 ^ { 2 } 	imes 0.5 ^ { 1 } 	imes 0.1 ^ { 2 }$$
Therefore, the answer is A.

Using the Formula for the Multinomial Distribution, the Probability Of X