In a survey of a group of people regarding their preference for two drinks, Tea and Coffee, it was found that 50 % of the people like Tea, 60 % like Coffee, and 90 % like at least one of the two drinks. If a person is selected at random, what is the sum of the probability that the person likes Tea given that they do not like Coffee, and the probability that the person likes Coffee given that they do not like Tea?

Step-by-step solution

Let T and C be the events that a randomly selected person likes Tea and Coffee, respectively. Given: P(T) = 0.5 P(C) = 0.6 P(T C) = 0.9 Using the inclusion-exclusion principle: P(T C) = P(T) + P(C) - P(T C) 0.9 = 0.5 + 0.6 - P(T C) P(T C) = 1.1 - 0.9 = 0.2 The probability of liking Tea but not Coffee is: P(T C') = P(T) - P(T C) = 0.5 - 0.2 = 0.3 The probability of liking Coffee but not Tea is: P(C T') = P(C) - P(T C) = 0.6 - 0.2 = 0.4 The probabilities of the complementary events are: P(C') = 1 - P(C) = 1 - 0.6 = 0