The probability that a randomly selected STAT 200 student is

The probability that a randomly selected STAT 200 student is a regular coffee drinker is 0.40, and the probability that the student is a regular tea drinker is 0.35. The probability that the student is both a regular coffee drinker and a regular tea drinker is 0.15. Find the probability that the student is neither a regular coffee nor a regular tea drinker. A) 0.14 B) 0.40 C) 0.60 D) 0.85 E) 0.75

Solution

Let A represents the regular coffee drinker and

B represents the regular tea drinker

=>

P(A) = 0.40

P(B) = 0.35

P(A ? B) = 0.15

we know that

percentage of perople drink either coffee or tea

= P( A U B) = P(A) + P(B) - P(A ? B) = 0.6

no of perople drink either tea or coffee = 0.6 * 200 = 120

no of people who is neither coffee or tea drinker = 200 -120 = 80

required probability = 80/200 = 0.40

hence B is the correct choice