Question

Suppose that 16% of people have a dog. 32% of people have a cat, and 10% of people own both. What is the probability that someone owns a dog or a cat?

Answer 1

0.38

Explanation:

Let the even that a person has a dog be A. The probability is given 16% this implies:

P(A) = 0.16 ..... (1)

Let the event that a person has a cat denote by B. The probability of B is given to be 32% this implies....

P(B) = 0.32 ..... (2)

Then the event that a person has a dog and a cat will be denoted by A∩B. The probability of A∩B is given to be 10%: This implies:

P(A∩B= = 0.12 .......(3)

The event that someone owns a cat and a dog will denoted by the event A∪B. It is observed that A and B are not disjoint. This implies that the probability of this event is depicted as:

P(A∪B) = P (A) + P(B) - P(A∩B)

Substitute (1) , (2) and (3), in the above, we get:

P(A∪B) = 0.16 + 0.32 - 0.10 = 0.38