Two fair dice are thrown. The probability that the sum is a prime number is
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Prime sums and their ways: 2 (1), 3 (2), 5 (4), 7 (6), 11 (2). Total 15 of 36, so \(\dfrac{15}{36} = \dfrac{5}{12}\).
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Two fair dice are thrown. The probability that the sum is a prime number is
Prime sums and their ways: 2 (1), 3 (2), 5 (4), 7 (6), 11 (2). Total 15 of 36, so \(\dfrac{15}{36} = \dfrac{5}{12}\).
\(A\) and \(B\) are independent with \(P(A) = \tfrac25\) and \(P(B) = \tfrac14\). Then \(P(A \cup B)\) is
\(P(A \cap B) = \tfrac25 \cdot \tfrac14 = \tfrac{1}{10}\). \(P(A \cup B) = \tfrac25 + \tfrac14 - \tfrac{1}{10} = \tfrac{8 + 5 - 2}{20} = \tfrac{11}{20}\).
Trap: Adding \(P(A) + P(B) = 13/20\) counts the overlap twice.
Find the variance of the data \(2, 4, 6, 8, 10\).
Mean 6. Squared deviations \(16, 4, 0, 4, 16\), total 40. Variance \(= \dfrac{40}{5} = 8\).
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