Chapter 10.1, Problem 28E

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 $P\left(A\right)=0.7,P\left(B\right)=0.4$ , and $P\left(A\text{ and }B\right)=\mathrm{0.2.}$

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:

  $P\left(A\text{ and }B\right)=0.2$

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

  $\begin{array}{l}P\left(B\right)-P\left(A\text{ and }B\right)=0.7-0.2\\ =0.5\end{array}$

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

  $\begin{array}{l}P\left(\text{neither}\right)=1-0.5-0.2-0.2\\ =0.1\end{array}$

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 $0.5$ .

Given information:

  $A$ and $B$ are events in a sample space $S$ such that $P\left(A\right)=0.7,P\left(B\right)=0.4$ , and $P\left(A\text{ and }B\right)=\mathrm{0.2.}$

Calculation:

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

  $\begin{array}{l}P\left(B\right)-P\left(A\text{ and }B\right)=0.7-0.2\\ =0.5\end{array}$

Hence, the probability that $A$ occurs but $B$ does not is $0.5$ .

(c).

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

The probability that neither $A$ nor $B$ occurs is $10\%$ .

Given information:

  $A$ and $B$ are events in a sample space $S$ such that $P\left(A\right)=0.7,P\left(B\right)=0.4$ , and $P\left(A\text{ and }B\right)=\mathrm{0.2.}$

Calculation:

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

  $\begin{array}{l}P\left(A\text{ or }B\right)=0.7+0.4-0.2\\ =1.1-0.2\\ =0.9\end{array}$

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

  $\begin{array}{l}P\left(\text{neither }A\text{ nor }B\right)=P\left(\text{not}\left(A\text{ or B}\right)\right)\\ =1-P\left(A\text{ or }B\right)\\ =1-0.9\\ =0.1\\ =10\%\end{array}$

Hence, The probability that neither $A$ nor $B$ occurs is $10\%$ .

(d).

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

The probability that $A$ or $B$ occurs is $0.1$ .

Given information:

  $A$ and $B$ are events in a sample space $S$ such that $P\left(A\right)=0.7,P\left(B\right)=0.4$ , and $P\left(A\text{ and }B\right)=\mathrm{0.2.}$

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

  $\begin{array}{l}P\left(\text{neither}\right)=1-0.5-0.2-0.2\\ =0.1\end{array}$

Hence, the probability that $A$ or $B$ occurs is $0.1$ .

(e).

To verify: Are events $A$ or $B$ independent?.

The events are not independent.

Given information:

  $A$ and $B$ are events in a sample space $S$ such that $P\left(A\right)=0.7,P\left(B\right)=0.4$ , and $P\left(A\text{ and }B\right)=\mathrm{0.2.}$

Verify:

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

  $\begin{array}{l}P\left(A\right)=0.7\\ P\left(A|B\right)=0.2\end{array}$

  $\begin{array}{l}P\left(A|B\right)=\frac{P\left(A\text{ and }B\right)}{P\left(B\right)}\\ =\frac{0.2}{0.4}=0.5\ne 0.7=P\left(A\right)\\ \end{array}$

  $P\left(A\right)\ne P\left(A|B\right)$

Hence, the events are not independent.