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A single die is rolled twice. Find the probability of rolling an odd number the first time and a number greater than 2 the second time. Find the probability of rolling an odd number the first time and a number greater than 2 the secon time. \( \square \) (Type an integer or a simplified fraction.)

Ask by Robbins Montgomery. in the United States
Mar 20,2025

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Answer

The probability is \(\frac{1}{3}\).

Solution

1. The probability of rolling an odd number on a single die is \[ \frac{3}{6} = \frac{1}{2}, \] because the odd numbers are \(1\), \(3\), and \(5\). 2. The probability of rolling a number greater than 2 is \[ \frac{4}{6} = \frac{2}{3}, \] because the numbers greater than 2 are \(3\), \(4\), \(5\), and \(6\). 3. Since the rolls are independent, the combined probability of both events is the product of the individual probabilities: \[ \frac{1}{2} \times \frac{2}{3} = \frac{2}{6} = \frac{1}{3}. \] Thus, the probability is \(\frac{1}{3}\).

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To solve this problem, let's break it down step by step. In the first roll, there are three odd numbers on a die: 1, 3, and 5. So, the probability of rolling an odd number first is: \[ P(\text{Odd on first roll}) = \frac{3}{6} = \frac{1}{2} \] In the second roll, the numbers greater than 2 are: 3, 4, 5, and 6. Therefore, the probability of rolling a number greater than 2 is: \[ P(\text{>2 on second roll}) = \frac{4}{6} = \frac{2}{3} \] Now, because the rolls are independent, we can multiply their probabilities together: \[ P(\text{Odd first and >2 second}) = P(\text{Odd first}) \times P(\text{>2 second}) = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3} \] So, the final probability is: \[ \frac{1}{3}. \]

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