To find the probability that a student chosen randomly from the junior class takes Algebra 2 or Chemistry, we can use the principle of inclusion-exclusion.
Let:
- \( A \) be the set of students taking Algebra 2.
- \( C \) be the set of students taking Chemistry.
From the problem, we know:
- Total number of students, \( N = 325 \)
- Number of students taking Algebra 2, \( |A| = 190 \)
- Number of students taking Chemistry, \( |C| = 165 \)
- Number of students taking neither, which we denote as \( |N| = 40 \)
First, we need to determine the number of students taking either Algebra 2 or Chemistry. Since 40 students take neither, the number of students taking at least one of the two subjects is:
\[ |A \cup C| = N - |N| = 325 - 40 = 285 \]
Next, we can find the probability \( P(A \cup C) \) that a randomly chosen student takes either Algebra 2 or Chemistry:
\[ P(A \cup C) = \frac{|A \cup C|}{N} = \frac{285}{325} \]
Now, let's simplify \( \frac{285}{325} \):
The greatest common divisor of 285 and 325 is 5.
Dividing both the numerator and the denominator by 5:
\[ \frac{285 \div 5}{325 \div 5} = \frac{57}{65} \]
Thus, the probability that a randomly chosen student takes Algebra 2 or Chemistry is:
\[ \frac{57}{65} \]
The answer is \( \frac{57}{65} \).
So, the correct option is:
57/65