Suppose we throw a die once. (i) What is the probability of getting a number greater than 4? (ii) What is the probability of getting a number less than or equal to 4?

Step 1: Identify all possible outcomes

When a standard six-sided die is thrown once, the possible outcomes are the integers from 1 to 6. This set of all possible outcomes is called the sample space. The total number of these outcomes is 6.

Step 2: Determine outcomes for event (i)

For the first part of the problem, we define event E as getting a number greater than 4. By examining our sample space, the numbers that satisfy this condition are 5 and 6. Therefore, there are 2 favorable outcomes for event E.

Step 3: Calculate probability for event (i)

The probability of an event is calculated by dividing the number of favorable outcomes by the total number of possible outcomes. Substituting the values we found, the probability of getting a number greater than 4 is $\frac{2}{6}$, which simplifies to $\frac{1}{3}$.

Step 4: Determine outcomes for event (ii)

For the second part, we define event F as getting a number less than or equal to 4. From the sample space, the numbers that meet this criterion are 1, 2, 3, and 4. Thus, there are 4 favorable outcomes for event F.

Step 5: Calculate probability for event (ii)

Using the formula for probability again, we divide the 4 favorable outcomes for event F by the total of 6 possible outcomes. This gives us a probability of $\frac{4}{6}$, which simplifies to $\frac{2}{3}$.

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