Question

At a sporting event, a concession stand that sells hot dogs has two options for condiments, ketchup and mustard. It is known that 77% of customers request ketchup on their hot dogs, 52% request mustard, and 44% request both. If a customer is selected at random, what is the probability that they prefer their hot dog to be plain, i.e., with neither ketchup nor mustard? Express your answer as a decimal rounded to two decimal places.

Answer 1

P(neither ketchup nor mustard) = 1 - P(either ketchup or mustard)

= 1 - (P(ketchup) + P(mustard) - P(ketchup and mustard))

= 1 - (0.77 + 0.52 - 0.44)

= 1 - 0.85 = 0.15