Chapter 5.4, Problem 1E

Use the multinomial formula and find the probabilities for each.

a. n = 6, X1 = 3, X2 = 2, X3 = 1, p1 = 0.5, p2 = 0.3, p3 = 0.2

b. n = 5, X1 = 1, X2 = 2, X3 = 2, p1 = 0.3, p2 = 0.6, p3 = 0.1

c. n = 4, X1 = 1, X2 = 1, X3 = 2, p1 = 0.8, p2 = 0.1, p3 = 0.1

To determine

To find: The probability, $P\left(X\right)$

Answer to Problem 1E

$P\left(X\right)=\underset{\_}{0.135}$

Explanation of Solution

Given info:

The given values are $n=6,X_1=3,X_2=2,X_3=1,p_1=0.5,p_2=0.3\text{ and }{p}_3=0.2$

Calculation:

The formula for multinomial distribution is given below:

$P\left(X\right)=\frac{n!}{X_1!X_2!X_3!\cdot \cdot \cdot X_k!}{p}_1^{X_1}{p}_2^{X_2}{p}_3^{X_3}\cdot \cdot \cdot p_k^{X_k}$

Where, $X_1+X_2+X_3+\cdot \cdot \cdot +X_k=n$ and $p_1+p_2+p_3+\mathrm{...}+p_k=1$

Substitute n as 6, $X_1$ as 3, $X_2$ as 2, $X_3$ as 1 and $p_1\text{ as }0.5,p_2\text{ as }0.3\text{ and }{p}_3\text{ as }0.2$

$\begin{array}{c}P\left(X\right)=\frac{6!}{3!2!1!}{\left(0.5\right)}^3{\left(0.3\right)}^2{\left(0.2\right)}^1\\ =\frac{720}{12}\left(0.125\right)\left(0.09\right)\left(0.2\right)\\ =60\left(0.125\right)\left(0.09\right)\left(0.2\right)\\ =0.135\end{array}$

To determine

To find: The probability, $P\left(X\right)$

Answer to Problem 1E

$P\left(X\right)=\underset{\_}{0.0324}$

Explanation of Solution

Given info:

The given values are $n=5,X_1=1,X_2=2,X_3=2,p_1=0.3,p_2=0.6\text{ and }{p}_3=0.1$

Calculation:

Substitute n as 5, $X_1$ as 1, $X_2$ as 2, $X_3$ as 2 and $p_1\text{ as }0.3,p_2\text{ as }0.6\text{ and }{p}_3\text{ as }0.1$

$\begin{array}{c}P\left(X\right)=\frac{5!}{1!2!2!}{\left(0.3\right)}^1{\left(0.6\right)}^2{\left(0.1\right)}^2\\ =\frac{120}{4}\left(0.3\right)\left(0.36\right)\left(0.01\right)\\ =30\left(0.00108\right)\\ =0.0324\end{array}$

To determine

To find: The probability, $P\left(X\right)$

Answer to Problem 1E

$P\left(X\right)=\underset{\_}{0.0096}$

Explanation of Solution

Given info:

The given values are $n=4,X_1=1,X_2=1,X_3=2,p_1=0.8,p_2=0.1\text{ and }{p}_3=0.1$

Calculation:

Substitute n as 4, $X_1$ as 1, $X_2$ as 1, $X_3$ as 2 and $p_1\text{ as }0.8,p_2\text{ as }0.1\text{ and }{p}_3\text{ as }0.1$

$\begin{array}{c}P\left(X\right)=\frac{4!}{1!1!2!}{\left(0.8\right)}^1{\left(0.1\right)}^1{\left(0.1\right)}^2\\ =\frac{24}{2}\left(0.8\right)\left(0.1\right)\left(0.01\right)\\ =12\left(0.0008\right)\\ =0.0096\end{array}$