Problem 52 - Algebra and Trigonometry

Question

A single die is rolled twice. Find the probability of rolling an odd number the first time and a number less than 3 the second time.

Step-by-Step Solution

Answer

The probability of rolling an odd number the first time and a number less than 3 the second time is \( \frac{1}{6} \).

Step 1: Determine the probability of rolling an odd number

An odd number is either 1, 3, or 5. So there are 3 favorable outcomes. Since a die has 6 faces, the total number of outcomes is 6. Probability is defined as the ratio of the number of favorable outcomes to the total number of outcomes. Hence, the probability of rolling an odd number \( P(\text{Odd}) \) can be calculated as follows: \( P(\text{Odd}) = \frac{\text{number of favorable outcomes}}{\text{total number of outcomes}} = \frac{3}{6} = \frac{1}{2} \).

Step 2: Determine the probability of rolling a number less than 3

Numbers less than 3 are: 1 and 2. Hence, the number of favorable outcomes in this case is 2. Again, since a die has 6 faces, the total number of outcomes is 6. Therefore, the probability of rolling a number less than 3 \( P(<3) \) is calculated as follows: \( P(<3) = \frac{\text{number of favorable outcomes}}{\text{total number of outcomes}} = \frac{2}{6} = \frac{1}{3} \).

Step 3: Determine the combined probability of both events

These are two independent events, the probability of both events occurring simultaneously is the product of their individual probabilities. Therefore, the required probability \( P(\text{Odd and} < 3) \) is calculated as follows: \( P(\text{Odd and} < 3) = P(\text{Odd}) \times P(<3) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} \).