To solve this problem, we will use the principle of inclusion-exclusion to find the number of female students who play either field hockey, volleyball, or both.
Let:
- \( A \) be the set of students who play field hockey.
- \( B \) be the set of students who play volleyball.
Given:
- \( |A| = 98 \) (students who play field hockey)
- \( |B| = 62 \) (students who play volleyball)
- \( |A \cap B| = 40 \) (students who play both sports)
We want to find \( |A \cup B| \), which is the number of students who play either field hockey or volleyball. According to the principle of inclusion-exclusion, we have:
\[ |A \cup B| = |A| + |B| - |A \cap B| \]
Substituting the numbers we have:
\[ |A \cup B| = 98 + 62 - 40 \]
Calculating this gives:
\[ |A \cup B| = 160 \]
Now, to find the probability that a female student selected at random plays either field hockey or volleyball, we need to calculate the ratio of students who play either sport to the total number of female students.
Let \( n = 200 \) (the total number of female students).
The probability \( P \) that a randomly selected student plays field hockey or volleyball is:
\[ P = \frac{|A \cup B|}{n} = \frac{160}{200} \]
Now we simplify this fraction:
\[ \frac{160}{200} = \frac{16}{20} = \frac{4}{5} \]
Thus, the probability that a female student selected at random plays either field hockey or volleyball is:
\[ \boxed{\frac{4}{5}} \]
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