Question:
Let $A$ and $B$ are two events and $P \left( A '\right)=0.3, P ( B )=0.4, P \left( A \cap B '\right)=0.5$, then $P ( A \cup
B)$ is.
Let $A$ and $B$ are two events and $P \left( A '\right)=0.3, P ( B )=0.4, P \left( A \cap B '\right)=0.5$, then $P ( A \cup
B)$ is.
Updated On: Jun 18, 2022
- 0.5
- 0.9
- 1
- 0.1
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The Correct Option is B
Solution and Explanation
Given :- $P \left( A '\right)=0.3, P ( B )=0.4, P \left( A \cap B'\right)=0.5$
$P ( A \cup B )= P ( B )+ P \left( A \cap B '\right)$
$= P ( A \cup B )=0.4+0.5$
$=0.9$
$P ( A \cup B )= P ( B )+ P \left( A \cap B '\right)$
$= P ( A \cup B )=0.4+0.5$
$=0.9$
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Concepts Used:
Bayes Theorem
Bayes’ Theorem is a part of the conditional probability that helps in finding the probability of an event, based on previous knowledge of conditions that might be related to that event.
Mathematically, Bayes’ Theorem is stated as:-
\(P(A|B)=\frac{P(B|A)P(A)}{P(B)}\)
where,
- Events A and B are mutually exhaustive events.
- P(A) and P(B) are the probabilities of events A and B, respectively.
- P(A|B) is the conditional probability of the happening of event A, given that event B has happened.
- P(B|A) is the conditional probability of the happening of event B, given that event A has already happened.
This formula confines well as long as there are only two events. However, Bayes’ Theorem is not confined to two events. Hence, for more events.