Chapter 8.6, Problem 75E

(a)

To determine

Whether $A$ and $B$ are mutually exclusive, and to draw a diagram to support answer.

Answer to Problem 75E

Thus for mutually exclusive event is

  $1.34$

Explanation of Solution

Given information:

Let $A$ and $B$ be two events from the same sample space such that $P\left(A\right)=0.76$ and $P\left(B\right)=0.58$ .

Calculation:

Consider two events $A$ and $B$ from the same sample space such that

  $P\left(A\right)=0.76$ and $P\left(B\right)=0.58$

Event $A$ and $B$ are exclusive if $A$ and $B$ have no sample point in common or if

  $A\cap B$ is empty.

If $A\cap B$ is empty, then

  $\begin{array}{c}P\left(A\cap B\right)=P\left(\varphi \right)\\ =0\end{array}$

Thus for mutually exclusive event is

  $\begin{array}{c}P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)\\ =0.76+0.58-0\\ =1.34\end{array}$

But the probability $P\left(A\cup B\right)$ is not possible since the probability of any event cannot be greater than one

Thus, the events $A$ and $B$ are not mutually exclusive since the sum of probability is greater than one

The diagram showing the events $A$ and $B$ is shown below.

  

(b)

To determine

Whether $A^{\prime }$ and $B^{\prime }$ are mutually exclusive, and to draw a diagram to support answer.

Answer to Problem 75E

The sum of probability of the complementary events $A^{\prime }$ and $B^{\prime }$ is

  $0.66$

Explanation of Solution

Given information:

Let $A$ and $B$ be two events from the same sample space such that $P\left(A\right)=0.76$ and $P\left(B\right)=0.58$ .

Calculation:

The formula of the complement of the event is

  $P\left(X^{\prime }\right)=1-P\left(X\right)$

Therefore, the complement of $P\left(A\right)$ and $P\left(B\right)$ is

  $\begin{array}{c}P\left(A^{\prime }\right)=1-P\left(A\right)\\ =1-0.76\\ =0.24\end{array}$

And

  $\begin{array}{c}P\left(B^{\prime }\right)=1-P\left(B\right)\\ =1-0.58\\ =0.42\end{array}$

Thus, the sum of probability of the complementary events $A^{\prime }$ and $B^{\prime }$ is

  $P\left(A^{\prime }\right)$ and $P\left(B^{\prime }\right)$

  $\begin{array}{l}=0.24+0.42\\ =0.66\end{array}$

Thus, the events $A^{\prime }$ and $B^{\prime }$ are mutually exclusive since the sum of probability is less than one

The diagram showing the events $A^{\prime }$ and $B^{\prime }$ is shown below

  

(c)

To determine

To find: the possible range of $P\left(A\cup B\right)$ .

Answer to Problem 75E

The range of $P\left(A\cup B\right)$ is $0.76\le P\left(A\cup B\right)\le 1$

Explanation of Solution

Given information:

Let $A$ and $B$ be two events from the same sample space such that $P\left(A\right)=0.76$ and $P\left(B\right)=0.58$ .

Calculation:

Note that the probability of an event $A$ or $B$ is always greater than the individual probabilities of the events $A$ and $B$

That is

  $P\left(A\cup B\right)\ge P\left(A\right)\text{and}P\left(A\cup B\right)\ge P\left(B\right)$

Thus,

  $P\left(A\cup B\right)\ge 0.76$

Since the probability of an event is always smaller than 1, therefore

  $P\left(A\cup B\right)\le 1$

Therefore, the range of $P\left(A\cup B\right)$ is $0.76\le P\left(A\cup B\right)\le 1$