An artist entered two paintings (Lighthouse and Old Cathedral) in a show. He feels that the probabilities are, 0.15, 0.18, and 0.11, respectively, that he will sell the Lighthouse, the Old Cathedral, or both. What is the probability that he will sell either of the two paintings?
Hint:
P(A∩B)=P(A)+P(B)-P(AUB)
To find the probability that he will sell either of the two paintings, we can use the formula for the probability of the union of two events.
Let's first define the events:
A = event of selling the Lighthouse painting
B = event of selling the Old Cathedral painting
We are given the individual probabilities:
P(A) = 0.15 (probability of selling the Lighthouse)
P(B) = 0.18 (probability of selling the Old Cathedral)
We are also given the probability of both paintings being sold:
P(A∩B) = 0.11 (probability of selling both paintings)
Now, we can use the formula:
P(AUB) = P(A) + P(B) - P(A∩B)
Substituting the given values:
P(AUB) = 0.15 + 0.18 - 0.11
Calculating the probabilities:
P(AUB) = 0.22
Therefore, the probability that he will sell either of the two paintings is 0.22.