Question

two hundred fifty - five california faculty members were surveyed regarding their plans for the upcoming weekend. the results were as follows: 118 plan to grade papers 116 plan to attend a football game 110 plan to do household chores 17 plan to grade papers and attend a football game 54 plan to grade papers and do household chores 44 plan to attend a football game and do household chores 11 plan to grade papers, attend a football game, and do household chores 1. how many planned to only grade papers? 2. how many planned to grade papers or do household chores? 3. how many did not plan to do any of these activities? 4. how many planned to do at lease two of these activities? 5. how many planned to attend a football game and do household chores, but did not have any plans to grade? 6. how many planned to attend a football game or grade papers, but did not plan to do household chores? 7. how many planned to do at most one of these activities?

Explanation:

Step1: Define sets

Let $A$ be the set of those who plan to grade papers, $B$ be the set of those who plan to attend a football - game, and $C$ be the set of those who plan to do household chores. Given $n(A)=118$, $n(B)=116$, $n(C)=110$, $n(A\cap B)=17$, $n(A\cap C)=54$, $n(B\cap C)=44$, $n(A\cap B\cap C)=11$, and the total number of faculty members $N = 255$.

Step2: Calculate the number of those who only grade papers

The number of those who only grade papers is $n(A)-n(A\cap B)-n(A\cap C)+n(A\cap B\cap C)$.
$n(\text{only }A)=n(A)-(n(A\cap B)+n(A\cap C)) + n(A\cap B\cap C)=118-(17 + 54)+11=68$.

Step3: Calculate $n(A\cup C)$

Using the formula $n(A\cup C)=n(A)+n(C)-n(A\cap C)$, we have $n(A\cup C)=118 + 110-54=174$.

Step4: Calculate the number of those who do none of the activities

First, calculate $n(A\cup B\cup C)$ using the formula $n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(A\cap C)-n(B\cap C)+n(A\cap B\cap C)$.
$n(A\cup B\cup C)=118+116 + 110-17-54-44 + 11=240$.
The number of those who do none of the activities is $N - n(A\cup B\cup C)=255 - 240=15$.

Step5: Calculate the number of those who do at least two activities

The number of those who do at least two activities is $n(A\cap B)+n(A\cap C)+n(B\cap C)-2n(A\cap B\cap C)$.
$n(\text{at least two})=17+54 + 44-2\times11=93$.

Step6: Calculate the number of those who plan to attend a football - game and do household chores but not grade papers

The number of those who plan to attend a football - game and do household chores but not grade papers is $n(B\cap C)-n(A\cap B\cap C)=44 - 11=33$.

Step7: Calculate the number of those who plan to attend a football - game or grade papers but not do household chores

First, find $n((A\cup B)\cap\overline{C})$.
$n(A\cup B)=n(A)+n(B)-n(A\cap B)=118+116 - 17=217$.
$n((A\cup B)\cap C)=n((A\cap C)\cup(B\cap C))=n(A\cap C)+n(B\cap C)-n(A\cap B\cap C)=54 + 44-11=87$.
$n((A\cup B)\cap\overline{C})=n(A\cup B)-n((A\cup B)\cap C)=217-87=130$.

Step8: Calculate the number of those who do at most one activity

The number of those who do at most one activity is $N - n(\text{at least two})=255 - 93=162$.

Answer:

1. 68
2. 174
3. 15
4. 93
5. 33
6. 130
7. 162