Question

Let AA and BB be events such that P(A∩B)=1/73P(A∩B)=1/73, P(A~)=68/73P(A~)=68/73, P(B)=21/73P(B)=21/73. What is the probability of A ∪BA ∪B, P(A∪B)P(A∪B)?

Answer 1

Answer:

$P(A \cup B) = \frac{5}{73} +\frac{21}{73} -\frac{1}{73}=\frac{25}{73}$

Step-by-step explanation:

Let A and B events. We have defined the probabilities for some events:

$P(A') =\frac{68}{73} , P(B) =\frac{21}{73} , P(A \cap B) =\frac{1}{73}$

Where A' represent the complement for the event A

The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: $P(A)+P(A') =1$

So for this case we can solve for P(A) like this:

$P(A) = 1-P(A') = 1-\frac{68}{73}=\frac{5}{73}$

And now we can find $P(A \cup B)$ using the total probability rul given by:

$P(A \cup B) = P(A)+P(B) -P(A \cap B)$

And if we replace the values given we got:

$P(A \cup B) = \frac{5}{73} +\frac{21}{73} -\frac{1}{73}=\frac{25}{73}$

And that would be the final answer.