Out of 50 students, 24 are taking art and 30 are taking music. Sixteen of the students are taking both art and music. If two different students from this group are chosen at random, find the probability that both students take art. Express your answer as fraction in simplest form. Use a Venn diagram.
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Step-by-step explanation:
To solve this problem, you can use a Venn diagram to visualize the information and calculate the probability. Let's denote:
- \(A\) as the set of students taking art,
- \(M\) as the set of students taking music.
Given:
- Total number of students, \(n(U) = 50\),
- Number of students taking art, \(n(A) = 24\),
- Number of students taking music, \(n(M) = 30\),
- Number of students taking both art and music, \(n(A \cap M) = 16\).
Using the principle of inclusion-exclusion, the number of students taking either art or music is:
\[n(A \cup M) = n(A) + n(M) - n(A \cap M) = 24 + 30 - 16 = 38.\]
Now, to find the probability that both students selected take art (\(P(A \cap A)\)), we'll calculate it as the ratio of the number of students taking both art to the total number of students:
\[P(A \cap A) = \frac{n(A \cap M)}{n(U)} = \frac{16}{50} = \frac{8}{25}.\]
So, the probability that both students selected take art is \(8/25\).