Question

David wants to rent a movie. He wants to watch either a comedy or a drama. The movie rental store has 18 comedies and dramas available for rent. Seven of the movies are comedies, and eleven of the movies are dramas. David has not seen two of the comedies, and he has not seen four of the dramas. If David selects a movie randomly, what is the probability the movie will be a comedy or a movie that he has not seen?

Answer 1

Let
A = number of ways to pick a movie David wants (either a comedy or a movie he hasn't seen)

B = number of movies total

There are 7 comedies and 4 dramas that David hasn't seen (note: the comedy movies that he hasn't seen are part of the "comedy movie" count, so there's no need to double count these). So there are 7+4 = 11 movies that David would want to rent. Therefore, A = 11.

There are 18 movies total for rent, so B = 18

Dividing the two values
A/B = 11/18 = 0.6111 approximately

The answer as a fraction is exactly 11/18
The answer in decimal form is roughly 0.6111
The answer in percent form is approximately 61.11%

Answer 2

$|\Omega|=18\\ |A|=7+4=11\\\\ P(A)=\dfrac{11}{18}\approx61\%$