From a set \(A = [3, \sqrt{2}, 2\sqrt{3}, \sqrt{9}, \sqrt{7}]\), a number is selected at random. Find the probability that is a rational number
The correct answer is B. \(\frac{2}{5}\)
\(A = {3, \sqrt{2}, 2\sqrt{3}, \sqrt{9}, \sqrt{7}}\)
n(A) = 5
Let the rational nos = R
n(R) = 2 (3, \(\sqrt{9}\))
P(R) = 2/5
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