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A penny is tossed and a number cube is rolled. Find each probability. 26. \( P \) (heads and 4 ) 27. \( P \) (tails and 1 ) 28. \( P \) (tails and odd number) 29. \( P \) (heads and a number greater than 2 ) 30. \( P \) (tails and a number less than or equal to 5 )

Ask by Cummings Chandler. in the United States
Mar 17,2025

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- \( P(\text{heads and } 4) = \frac{1}{12} \) - \( P(\text{tails and } 1) = \frac{1}{12} \) - \( P(\text{tails and odd number}) = \frac{1}{4} \) - \( P(\text{heads and a number greater than } 2) = \frac{1}{3} \) - \( P(\text{tails and a number less than or equal to } 5) = \frac{5}{12} \)

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To solve for the probability of each event, we will consider the outcomes of tossing a penny and rolling a number cube (with faces numbered 1 through 6). A penny has 2 outcomes (heads or tails), and a number cube has 6 outcomes. Therefore, the total number of outcomes for this combined experiment is \( 2 \times 6 = 12 \). 26. \( P \) (heads and 4): There is 1 favorable outcome (Heads and 4), so \( P = \frac{1}{12} \). 27. \( P \) (tails and 1): There is 1 favorable outcome (Tails and 1), so \( P = \frac{1}{12} \). 28. \( P \) (tails and odd number): The odd numbers on the cube are 1, 3, and 5, resulting in 3 favorable outcomes (Tails with 1, 3, or 5), so \( P = \frac{3}{12} = \frac{1}{4} \). 29. \( P \) (heads and a number greater than 2): The numbers greater than 2 are 3, 4, 5, and 6, resulting in 4 favorable outcomes (Heads with 3, 4, 5, or 6), so \( P = \frac{4}{12} = \frac{1}{3} \). 30. \( P \) (tails and a number less than or equal to 5): The numbers less than or equal to 5 are 1, 2, 3, 4, and 5, resulting in 5 favorable outcomes (Tails with 1, 2, 3, 4, or 5), so \( P = \frac{5}{12} \).

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