Chapter 15, Problem 11CT

(a)

To determine

To find: The values of $P\left(A\right).P\left(B\right),P\left(A\cap B\right),P\left(A\cup B\right)$ .

Answer to Problem 11CT

The value of function is $P\left(A\cup B\right)=\frac{5}{9}$ .

Explanation of Solution

Given information:

There are $3$ red, $3$ white and $3$ green marble.

Calculation:

Calculate the expression

  $\begin{array}{c}P\left(A\right)=\frac{3}{9}\\ =\frac{1}{3}\\ P\left(B\right)=\frac{3}{9}\\ =\frac{1}{3}\\ P\left(A\cap B\right)=\frac{1}{9}\end{array}$

Two events $AandB$

  $\begin{array}{c}P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)\\ P\left(A\cup B\right)=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\\ P\left(A\cup B\right)=\frac{5}{9}\end{array}$

Therefore, the value of function is $P\left(A\cup B\right)=\frac{5}{9}$ .

(b)

To determine

To find: The independent.

Answer to Problem 11CT

The independent are $P\left(A\right)P\left(B\right)=P\left(A\cap B\right)$ .

Explanation of Solution

Given information:

The given expression $3$ red $3$ white and $3$ green marble

Calculation:

Calculate the expression.

  $P\left(A\cap B\right)=P\left(A\right)P\left(B\right)$

Calculate $P\left(A\right)P\left(B\right)$ .

  $\begin{array}{c}P\left(A\right)P\left(B\right)=\frac{1}{3}\left(\frac{1}{3}\right)\\ =\frac{1}{9}\end{array}$

Therefore, the independent events are $P\left(A\right)P\left(B\right)=P\left(A\cap B\right)$