Question

The theater director offered every member of the drama one vote for which play they preferred to perform. The director found that 3 5 % 35% voted for The Oddems Family, that 5 7 % 57% voted for Thirteenth Night, and that 8 % 8% did not vote. In this group, what is the probability this group voted for "Oddems Family" or "Thirteenth Night"

Answer 1

Answer:

0.92 is the required probability.

Step-by-step explanation:

We are given the following in the question:

Percentage who voted for The Oddems Family = 35%

$P(A) = 0.35$

Percentage who voted for The Thirteenth Night = 57%

$P(B) = 0.57$

Percentage who did not vote = 8%

$P(A^C\cap B^C) = 0.08$

We have to find the probability that the group voted for "Oddems Family" or "Thirteenth Night".

Thus, we have to evaluate

$P(A\cup B)$

This can be done in the following manner:

$P(A^C\cap B^C) = 0.08\\P(A\cup B)^C = 0.08\\P(A\cup B) = 1-0.08\\P(A\cup B) = 0.92$

Thus, 0.92 is the probability  that the group voted for "Oddems Family" or "Thirteenth Night".