Suppose 100 stastisticians attended a conference of the American Statistical Society. At the dinner, among the menu options were a Caesar salad, roast beef, and apple pie. 35 had the Caesar salad, 28 had the roast beef, and 45 had the apple pie for dessert. Also, 15 had at least two of those three offerings, and 2 had all three. How many attendees had none of the three

26 attendees had none of the three.

Step-by-step explanation:

The Venn diagram illustrating the situation is shown in the attached photo.

C represents the set of statisticians that had Caesar salad.

R represents the set of statisticians that had roast beef.

A represents the set of statisticians that had apple pie for dessert.

x represents the number that had Caesar salad and apple pie for dessert only.

y represents the number that had Caesar salad and roast beef.

z represents the number that had roast beef and apple pie for dessert only.

If 15 had at least two of those three offerings,it means that

x + y + z = 15

Therefore,

35 - (x + y + 2) + 28 - (y + z + 2) + 45 - (x + z + 2) + 2 + none = 100

35 - x - y - 2 + 28 - y - z - 2 + 45 - x - z - 2 + 2 + none = 100

35 + 28 + 45 - x - x - y - y - z - z - 2 - 2 - 2 + 2 + none = 100

108 - 2x - 2y - 2z - 4 + none = 100

108 - 4 - 2(x + y + z) + none = 100

Since x + y + z = 15, then

104 - 2(15) + none = 100

74 + none = 100

none = 100 - 74 = 26

is there more to this question?