# If P\left(A\right)=0.4, P\left(B\right)=0.5, and P\left(A\cap B\

If $$\displaystyle{P}{\left({A}\right)}={0.4}$$, $$\displaystyle{P}{\left({B}\right)}={0.5}$$, and $$\displaystyle{P}{\left({A}\cap{B}\right)}={0.3}$$, find a. $$\displaystyle{P}{\left({A}\cup{B}\right)}$$, b. $$\displaystyle{P}{\left({A}\cap{B}'\right)}$$, and c. $$\displaystyle{P}{\left({A}'\cup{B}'\right)}$$.

• Questions are typically answered in as fast as 30 minutes

### Plainmath recommends

• Get a detailed answer even on the hardest topics.
• Ask an expert for a step-by-step guidance to learn to do it yourself.

James Kilian

Explanation 1
Step 1
PROBABILITY RULES
Complement rule
$$\displaystyle{P}{\left({A}'\right)}={P}{\left(\neg{A}\right)}={1}-{P}{\left({A}\right)}$$
General addition rule for any two events:
$$\displaystyle{P}{\left({A}\cup{B}\right)}={P}{\left({A}\right)}+{P}{\left({B}\right)}-{P}{\left({A}\cap{B}\right)}$$
Step 2
SOLUTION
$$\displaystyle{P}{\left({A}\right)}={0.4}$$
$$\displaystyle{P}{\left({B}\right)}={0.5}$$
$$\displaystyle{P}{\left({A}\cap{B}\right)}={0.3}$$
(a) Use the general addition rule for any two events:
$$\displaystyle{P}{\left({A}\cup{B}\right)}={P}{\left({A}\right)}+{P}{\left({B}\right)}-{P}{\left({A}\cap{B}\right)}$$
$$\displaystyle={0.4}+{0.5}-{0.3}$$
$$\displaystyle={0.6}$$
Explanation 2
Step 1
Probability rules, used n the exercise:
Complement Rule:
$$\displaystyle{\left({A}'\right)}={P}{\left(\neg{A}\right)}={1}-{P}{\left({A}\right)}$$
$$\displaystyle{\left({A}\cup{B}\right)}={P}{\left({A}\right)}+{P}{\left({B}\right)}-{P}{\left({A}\cap{B}\right)}$$
Rule of Complementary events B and B':
$$\displaystyle{P}{\left({A}\right)}={P}{\left({A}\cap{B}\right)}+{P}{\left({A}\cap{B}'\right)}$$
De Morgan's Rules
$$\displaystyle{P}{\left({\left({A}\cup{B}\right)}'\right)}={P}{\left({A}'\cap{B}'\right)}$$
$$\displaystyle{P}{\left({\left({A}\cap{B}\right)}'\right)}={P}{\left({A}'\cap{B}'\right)}$$
Step 2
Solution:
$$\displaystyle{P}{\left({A}\right)}={0.4}$$
$$\displaystyle{P}{\left({B}\right)}={0.5}$$
$$\displaystyle{P}{\left({A}\cap{B}\right)}={0.3}$$
(a)
$$\begin{matrix} \\ &{\left(Addition Rule\right) } \\P \left(A \cup B\right) & {=} & P\left(A\right)-P\left(A \cap B\right)\end{matrix}$$
$$\displaystyle={0.4}+{0.5}-{0.3}$$
$$\displaystyle={0.6}$$

###### Have a similar question?
Crom1970

Explanation 1
Step 3
(b) The event of A occurring but not B is the event A without all outcomes that occur for A and B have in common.
$$\displaystyle{P}{\left({A}\cap{B}'\right)}={P}{\left({A}\right)}–{P}{\left({A}\cap{B}\right)}={0.4}-{0.3}={0.1}$$
Explanation 2
(b)
$$\begin{matrix} \\ &{\left(Rule\ of\ Complementary\ events \right) } \\P \left(A \cap B'\right) & {=} & P\left(A\right)-P\left(A \cap B\right)\end{matrix}$$
$$\displaystyle{0.4}-{0.3}$$
$$\displaystyle={0.1}$$

Don Sumner

Explanation 1
Step 4
(c) The event of A not occurring or B not occurring is the event that A and B cannot occur together.
Use the complement rule:
$$P\left(A'\cup B'\right)= P\left(\left(A \cap B\right)'\right)=1- P\left(A \cap B\right)=1-0.3 = 0.7$$
Explanation 2
(c)
$$\begin{matrix} \\ &{\left(De Morgan's Rule (2) \right) } \\P \left(A' \cup B'\right) & {=} & P\left(\left(A \cap B\right)'\right)\\ &{\left(Complement Rule \right) } & \\ & = & 1-P\left(A \cap B\right) \end{matrix}$$
$$=1-0.3$$
$$=0.7$$