Chapter 10.1, Problem 27E

Solution

(a).

To graph : The Venn diagram showing the overlapping sets $A$ and $B$ , and fill in the properties of the four regions formed.

Given information :

  $A$ and $B$ are events in a sample space $S$ such that $P\left(A\right)=0.6,P\left(B\right)=0.5$ , and $P\left(A\text{ and }B\right)=\mathrm{0.3.}$

Graph :

To draw the diagram, start by drawing two overlapping circles and labeling them $A$ and $B$ .

  

Interpretation :

Write in the probability of both events occurring in the area where the circles overlap:

  $P\left(A\text{ and }B\right)=0.3$

The probability of A alone occurring, substitute $P\left(A\text{ and }B\right)=0.3$ from $P\left(B\right)$

  $\begin{array}{l}P\left(B\right)-P\left(A\text{ and }B\right)=0.5-0.3\\ =0.2\end{array}$

The probability of neither of neither of the events occurring, subtract all the probabilities written in the circles from $1:$

  $\begin{array}{l}P\left(\text{neither}\right)=1-0.3-0.3-0.2\\ =0.2\end{array}$

Now, write this number outside of the circles and then draw the graph.

(b).

To calculate: The probability that $A$ occurs but $B$ does not.

The probability that $A$ occurs but $B$ does not is $\mathrm{0.3.}$

Given information:

  $A$ and $B$ are events in a sample space $S$ such that $P\left(A\right)=0.6,P\left(B\right)=0.5$ , and $P\left(A\text{ and }B\right)=\mathrm{0.3.}$

Calculation:

To find the probability $A$ alone occurring, subtract $P\left(A\text{ and }B\right)=0.3$ from $P\left(B\right)$

  $\begin{array}{l}P\left(B\right)-P\left(A\text{ and }B\right)=0.6-0.3\\ =0.3\end{array}$

(c).

To calculate: The probability that neither $A$ nor $B$ occurs.

The probability that neither $A$ nor $B$ occurs is $20\%$ .

Given information:

  $A$ and $B$ are events in a sample space $S$ such that $P\left(A\right)=0.6,P\left(B\right)=0.5$ , and $P\left(A\text{ and }B\right)=\mathrm{0.3.}$

Calculation:

Use the general addition rule for any two events: $P\left(A\text{ or }B\right)=P\left(A\right)+P\left(B\right)-P\left(A\text{ and }B\right)$

  $\begin{array}{l}P\left(A\text{ or }B\right)=0.6+0.5-0.3\\ =1.1-0.3\\ =0.8\end{array}$

Use the complementary rule: $P\left(\text{not }A\right)=1-P\left(A\right)$

  $\begin{array}{l}P\left(\text{neither }A\text{ nor }B\right)=P\left(\text{not}\left(A\text{ or B}\right)\right)\\ =1-P\left(A\text{ or }B\right)\\ =1-0.8\\ =0.2\\ =20\%\end{array}$

Hence, the probability that neither $A$ nor $B$ occurs is $20\%$ .

(d).

To calculate: The probability that $A$ or $B$ occurs.

The probability that $A$ or $B$ occurs is $0.2$ .

Given information:

  $A$ and $B$ are events in a sample space $S$ such that $P\left(A\right)=0.6,P\left(B\right)=0.5$ , and $P\left(A\text{ and }B\right)=\mathrm{0.3.}$

Calculation:

To find the probability of neither of the events occurring, subtract all the probabilities written in the circles of the diagram from $1:$

Use the complementary rule: $P\left(\text{not }A\right)=1-P\left(A\right)$

  $\begin{array}{l}P\left(A\text{ or }B\text{ occurs}\right)=1-0.3-0.3-0.2\\ =0.2\end{array}$

Hence, the probability that $A$ or $B$ occurs is $0.2$ .

(e).

To calculate: The probability that $A$ or $B$ occurs.

The probability that $A$ or $B$ occurs is $0.2$ .

Given information:

  $A$ and $B$ are events in a sample space $S$ such that $P\left(A\right)=0.6,P\left(B\right)=0.5$ , and $P\left(A\text{ and }B\right)=\mathrm{0.3.}$

Verify:

We compare $P\left(A\right)$ and $\left(A|B\right)$ :

  $\begin{array}{l}P\left(A\right)=0.6\\ P\left(A|B\right)=0.3\\ \Rightarrow P\left(A\right)\ne P\left(A|B\right)\end{array}$

Hence, the events are not independent.