Question

1. A search using the Web search engine Bing^TM for "asteroid" yielded 24.8 million Web sites containing that word. A search for "comet" yielded 95.0 million sites. A search for sites containing both words yielded 3.8 million sites. How many Web sites contained either "asteroid" or "comet" or both? HINT [See Example 1.]
million sites

2. On a particularly boring transatlantic flight, one of the authors amused himself by counting the heads of the people in the seats in front of him. He noticed that all 26 of them either had black hair or had a whole row to themselves (or both). Of this total, 23 had black hair and 8 were fortunate enough to have a whole row of seats to themselves. How many of the black-haired people had whole rows to themselves?
people

Answer 1

(1) let A denote the set of "asteroid" and C denote the set of "comet"

Number of elements in set "asteroid" =n(A) = 24.8

number of elements in set 'comet" = n(C) = 95

number of elements common in both sets = n(A and C) = 3.8

we know the formula

n(A or C) = n(A) + n(C) - n(A and C).....................where n(A or C) is the required number of elements in set "asteroid" or "comet" or both.

setting the values, we get

n(A or C) = 24.8 + 95 - 3.8 = 119.8-3.8 = 116 million sites

Therefore, required answer is 116 million sites

(2) let B denote the set of black haired people and W denote the set of whole row

Number of elements in set "black haired" =n(B) = 23

number of elements in set "whole row" = n(W) = 8

number of elements who either had black hair or had a whole row to themselves (or both) = n(B or W) = 26

we know the formula

n(B and W) = n(B) + n(W) - n(B or W).....................where n(B and W) is the required number of elements in set black-haired people had whole rows to themselves

setting the values, we get

n(B and W) = 23 + 8 - 26 = 31-26 = 5 people

Therefore, required answer is 5 people