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

Jessica Scott 2021-11-20 Answered
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. \(\displaystyle{P}{\left({A}'\cup{B}'\right)}\).

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Expert Answer

James Kilian
Answered 2021-11-21 Author has 774 answers

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:
\(\displaystyle{\left({A}'\right)}={P}{\left(\neg{A}\right)}={1}-{P}{\left({A}\right)}\)
Addition Rule of general events:
\(\displaystyle{\left({A}\cup{B}\right)}={P}{\left({A}\right)}+{P}{\left({B}\right)}-{P}{\left({A}\cap{B}\right)}\)
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)
\(\begin{matrix} \\ &{\left(Addition Rule\right) } \\P \left(A \cup B\right) & {=} & P\left(A\right)-P\left(A \cap B\right)\end{matrix}\)
\(\displaystyle={0.4}+{0.5}-{0.3}\)
\(\displaystyle={0.6}\)

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Crom1970
Answered 2021-11-22 Author has 1197 answers

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)
\(\begin{matrix} \\ &{\left(Rule\ of\ Complementary\ events \right) } \\P \left(A \cap B'\right) & {=} & P\left(A\right)-P\left(A \cap B\right)\end{matrix}\)
\(\displaystyle{0.4}-{0.3}\)
\(\displaystyle={0.1}\)

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Don Sumner
Answered 2021-11-23 Author has 1842 answers

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)
\(\begin{matrix} \\ &{\left(De Morgan's Rule (2) \right) } \\P \left(A' \cup B'\right) &  {=} & P\left(\left(A \cap B\right)'\right)\\ &{\left(Complement Rule \right) } & \\ & = & 1-P\left(A \cap B\right) \end{matrix}\)
\(=1-0.3\)
\(=0.7\)

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