91Ó°ÊÓ

Let's represent the probability of rolling 1 or 2 as \(3x\), and the probability of rolling any other number as \(x\). The total probability of all six numbers should be equal to 1. We have a total of six numbers on the die: two numbers with a probability of \(3x\) and four with a probability of \(x\), so the total probability is \(2(3x) + 4(x)\), which should be equal to 1. So, we have: \(2(3x) + 4(x) = 1\) Now, we will solve for \(x\). \(6x + 4x = 1\) \(10x = 1\) \(x = \frac{1}{10}\) Now that we know the value of \(x\), we can find the probabilities of each number. Probabilities: - Rolling a 1 or 2: \(3x = 3\left(\frac{1}{10}\right) = \frac{3}{10}\) - Rolling a 3, 4, 5, or 6: \(x = \frac{1}{10}\) The probability distribution of this weighted die is: - P(1) = P(2) = \(\frac{3}{10}\) - P(3) = P(4) = P(5) = P(6) = \(\frac{1}{10}\)

Now, we will find the probability of rolling an even number (2, 4, or 6). To find the probability, we add up the probabilities of rolling a 2, 4, or 6: P(even) = P(2) + P(4) + P(6) = \(\frac{3}{10} + \frac{1}{10} + \frac{1}{10} = \frac{5}{10} = \frac{1}{2}\) So, the probability of rolling an even number on this weighted die is \(\frac{1}{2}\), or 50%.