Answer to Question #224315 in Statistics and Probability for Harry

Question

A sample of 200 tourists in Florida shows 52 went to Disney, 57 went to Busch Gardens, and 52 visited both. Find the probability that a tourist attended neither Disney nor Busch Gardens.

Answer 1

$$P(D\cup BG)=P(D)+P(BG)-P(D\cap BG)$$

$$=\dfrac{52}{200}+\dfrac{57}{200}-\dfrac{52}{200}=0.285$$

$$1-P(D\cup BG)=1-0.285=0.715$$

The probability that a tourist attended neither Disney nor Busch Gardens is $0.715.$