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
1 answer:
Answer: 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
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