Chapter 5.4, Problem 2E

Use the multinomial formula and find the probabilities for each.

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

b. n = 5, X₁ = 1, X₂ = 3, X₃ = 1, p₁ = 0.7, p₂ = 0.2, p₃ = 0.1

c. n = 7, X₁ = 2, X₂ = 3, X₃ = 2, p₁ = 0.4, p₂ = 0.5, p₃ = 0.1

To determine

The probability, $P\left(X\right)$

Answer to Problem 2E

$P\left(X\right)=\underset{\_}{0.18}$

Explanation of Solution

Given info:

When the values are $n=3,X_1=1,X_2=1,X_3=1,p_1=0.5,p_2=0.3\text{ and }{p}_3=0.2$

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, $X_1+X_1+\mathrm{...}+X_k=1$ and $p_1+p_2+\mathrm{...}+p_k=1$

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

$\begin{array}{c}P\left(X\right)=\frac{3!}{1!1!1!}{\left(0.5\right)}^1{\left(0.3\right)}^1{\left(0.2\right)}^1\\ =6\left(0.5\right)\left(0.3\right)\left(0.2\right)\\ =0.18\end{array}$

To determine

The probability, $P\left(X\right)$

Answer to Problem 2E

$P\left(X\right)=\underset{\_}{0.112}$

Explanation of Solution

Given info:

When the values are $n=5,X_1=1,X_2=3,X_3=1,p_1=0.7,p_2=0.2\text{ and }{p}_3=0.1$

Calculation:

Substitute $n\text{ as }5,X_1\text{ as }1,X_2\text{ as }3,X_3\text{ as }1,p_1\text{ as }0.7,p_2\text{ as }0.2\text{ and }{p}_3\text{ as }0.1$

$\begin{array}{c}P\left(X\right)=\frac{5!}{1!3!1!}{\left(0.7\right)}^1{\left(0.2\right)}^3{\left(0.1\right)}^1\\ =\frac{120}{6}\left(0.7\right)\left(0.008\right)\left(0.1\right)\\ =20\left(0.7\right)\left(0.008\right)\left(0.1\right)\\ =0.112\end{array}$

To determine

The probability, $P\left(X\right)$

Answer to Problem 2E

$P\left(X\right)=\underset{\_}{0.036}$

Explanation of Solution

Given info:

When the values are $n=7,X_1=2,X_2=3,X_3=2,p_1=0.4,p_2=0.5\text{ and }{p}_3=0.1$

Calculation:

Substitute $n\text{ as }7,X_1\text{ as }2,X_2\text{ as }2,X_3\text{ as }2,p_1\text{ as }0.4,p_2\text{ as }0.5\text{ and }{p}_3\text{ as }0.1$

$\begin{array}{c}P\left(X\right)=\frac{7!}{2!3!2!}{\left(0.4\right)}^2{\left(0.3\right)}^3{\left(0.1\right)}^2\\ =\frac{5,040}{6}\left(0.16\right)\left(0.027\right)\left(0.01\right)\\ =840\left(0.0000432\right)\\ =0.036\end{array}$