Answer:
P(O∪T)= 0.92
Step-by-step explanation:
Hello!
<em>The theater director offered every member of the drama one vote for which play they preferred to perform. The director found that 35% voted for "The Oddems Family", that 57% voted for "Thirteenth Night" and 8% did not vote. </em>
You have three events with three associated probabilities:
O: The performer voted for Oddems Family. ⇒ P(O)= 0.35
T: The performer voted for the Thirteenth Night. ⇒ P(T)= 0.57
n: The performer did not vote. ⇒ P(N)= 0.08
<em>In this group, are the events "Oddems Family" and "Thirteenth Night" mutually exclusive? </em>
Two events are mutually exclusive when the occurrence of one of them prevents the occurrence of the other in a single repeat of the trial, i.e. these two events cannot occur at the same time.
In this case, each performer has only one vote and can decide for either one of the plays, so if he votes for one, then he'll not vote for the other one. Then we can say that the events are mutually exclusive.
<em>Find the probability that a randomly selected person from this group voted for "Oddems Family" </em><u><em>or</em></u><em> "Thirteenth Night"</em>
You need to calculate the probability of a randomly selected performer to vote for either "Oddems Family" or "Thirteenth Night", this is a union between the two events and you can symbolize it as:
P(O∪T)
Remember:
Given the events A and B, that are mutually exclusive, you can calculate the union between them as:
P(A∪B)= P(A) + P(B)
Applying this to the exercise:
P(O∪T)= P(O) + P(T)= 0.35 + 0.57 = 0.92
I hope this helps!
