the associate head of a certain business school is looking to hire 3 undergraduate students as teaching assistants for the introduction to business analytics course. there is currently a pool of 12 students consisting of 4 finance majors, 6 accounting majors, and 2 marketing majors. believing all students are equally capable, the associate head will randomly pick 3 students (without replacement) from the pool of 12 teaching a candidate candidates. x is the random variable that counts the number of accounting majors that are to be hired as teaching assistants. complete the probability distribution table for x below. use four decimals in your probability computations.

Question

Accepted Answer

# Explanation:
## Step1: Calculate $P(X = 1)$
Use the hypergeometric formula $P(X=k)=\frac{\binom{K}{k}\binom{N - K}{n - k}}{\binom{N}{n}}$, where $N = 12$ (total students), $K = 6$ (Accounting majors), $n=3$ (students to be picked).
For $k = 1$:
$$
\begin{align*}
P(X = 1)&=\frac{\binom{6}{1}\binom{12- 6}{3 - 1}}{\binom{12}{3}}\
&=\frac{6\times\binom{6}{2}}{\frac{12!}{3!(12 - 3)!}}\
&=\frac{6\times\frac{6!}{2!(6 - 2)!}}{\frac{12\times11\times10}{3\times2\times1}}\
&=\frac{6\times15}{220}\
&=\frac{90}{220}\approx0.4091
\end{align*}
$$
## Step2: Calculate $P(X = 2)$
For $k = 2$:
$$
\begin{align*}
P(X = 2)&=\frac{\binom{6}{2}\binom{12-6}{3 - 2}}{\binom{12}{3}}\
&=\frac{\frac{6!}{2!(6 - 2)!}\times6}{\frac{12!}{3!(12 - 3)!}}\
&=\frac{15\times6}{220}\
&=\frac{90}{220}\approx0.4091
\end{align*}
$$

# Answer:
The values of $x$ are $0$, $1$, $2$, $3$.
$P(X = 1)=0.4091$, $P(X = 2)=0.4091$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step1: Calculate \(P(X = 1)\)

Step2: Calculate \(P(X = 2)\)

Answer: