In a survey of a group of people regarding their preferences for three different fruits—apples, bananas, and cherries—it was found that 50 people like apples, 45 like bananas, and 40 like cherries. If exactly 60 people like exactly one of these fruits, and 5 people like all three fruits, then what is the total number of people who like at least one of the three fruits?

Step-by-step solution

Let A , B , and C denote the sets of people who like apples, bananas, and cherries, respectively. Given: n(A) = 50 , n(B) = 45 , n(C) = 40 Number of people liking exactly one fruit, E₁ = 60 Number of people liking all three fruits, E₃ = 5 The sum of the individual set sizes is: S₁ = n(A) + n(B) + n(C) = 50 + 45 + 40 = 135 We know the relation between S₁ and the exact occurrence counts E₁ , E₂ (exactly two), and E₃ (exactly three): S₁ = E₁ + 2E₂ + 3E₃ Substituting the known values: 135 = 60 + 2E₂ + 3(5) 135 = 75 + 2