To solve the problem of finding the probability that the first roll of a six-sided number cube is an even number and the second roll is a number greater than 4, let's break it down step by step.
First, we need to determine the probability that the first roll is an even number. On a six-sided number cube (a standard die), the numbers are 1, 2, 3, 4, 5, and 6. The even numbers among these are 2, 4, and 6. Therefore, there are 3 even numbers out of the 6 possible outcomes.
The probability of rolling an even number on the first roll is:
[tex]\[ P(\text{even first roll}) = \frac{\text{number of even numbers}}{\text{total possible outcomes}} = \frac{3}{6} = \frac{1}{2} \][/tex]
Next, we need to determine the probability that the second roll is a number greater than 4. The numbers greater than 4 on a six-sided die are 5 and 6. Therefore, there are 2 numbers greater than 4 out of the 6 possible outcomes.
The probability of rolling a number greater than 4 on the second roll is:
[tex]\[ P(\text{greater than 4 second roll}) = \frac{\text{number of numbers greater than 4}}{\text{total possible outcomes}} = \frac{2}{6} = \frac{1}{3} \][/tex]
Since the rolls of a die are independent of each other, the combined probability of both events happening (the first roll being an even number and the second roll being a number greater than 4) is the product of the individual probabilities:
[tex]\[ P(\text{even first roll and greater than 4 second roll}) = P(\text{even first roll}) \times P(\text{greater than 4 second roll}) \][/tex]
Substituting the probabilities we calculated:
[tex]\[ P(\text{even first roll and greater than 4 second roll}) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} \][/tex]
So, the probability that the first roll is an even number and the second roll is a number greater than 4 is [tex]\(\frac{1}{6}\)[/tex].
Therefore, the correct answer is:
[tex]\(\boxed{\frac{1}{6}}\)[/tex]
