To solve the problem, we need to determine [tex]\( P(A^C \cup B^C) \)[/tex], which is the probability that a student does not take statistics or is not a senior. Here's a step-by-step breakdown of the solution:
1. Determine the total number of students surveyed:
The total number of students surveyed is given as 100.
2. Calculate the number of students who take statistics:
- Seniors taking statistics: 15
- Juniors taking statistics: 18
Total students taking statistics:
[tex]\[
15 + 18 = 33
\][/tex]
3. Calculate the number of students who take calculus:
- Seniors taking calculus: 35
- Juniors taking calculus: 32
Total students taking calculus:
[tex]\[
35 + 32 = 67
\][/tex]
4. Calculate the number of seniors:
- Seniors taking statistics: 15
- Seniors taking calculus: 35
Since no information suggests overlap within seniors taking both subjects, we assume the seniors' total is:
[tex]\[
15 + 35 = 50
\][/tex]
However, note that we need only the count of seniors who are involved in both statistics and calculus in the denominator.
5. Calculate the number of students taking either statistics or calculus:
Since 33 students are taking statistics and 67 are taking calculus, if there’s no overlap, the same has to involve just simple addition:
[tex]\[
33 + 67 - 33 = 67
\][/tex]
So, we consider as:
[tex]\[
33 + 67 - 33 = 67.
\][/tex]
6. Calculate [tex]\( P(A^C \cup B^C) \)[/tex]:
- [tex]\( P(A^C \cup B^C) = 1 - P(A \cap B) \)[/tex]
Where [tex]\( P(A \cap B) \)[/tex] is the probability that students take both statistics and calculus.
With 33 students taking statistics total, some of them also take calculus as:
[tex]\[
Total - Exclusive= 100 - 33 = 67
\][/tex]
So
[tex]\[
P(A^C \cup B^C) = 1 - \left(\frac{total_students - total_stats_or_calc}{total_students}\right) =\ 1 - (\frac{100 - 67}{100}) =\ \ 1- \left(\frac{33}{100}\right)
\][/tex]
Therefore:
[tex]\[
P(A^C \cup B^C )= 1 - 0.33 = 0.67
\][/tex]
So,
The final value, rounded to two decimal places, is:
[tex]\[
\boxed{0.67}
\][/tex]
