A die is tossed twice. The probability of having a number greater than 4 on each toss is

To solve this problem, we need to understand the basics of probability.

A die has 6 faces, numbered from 1 to 6. When we say "a number greater than 4", we are referring to the numbers 5 and 6. So, there are 2 favorable outcomes for each toss of the die.

Step 1: Calculate the probability of one event The probability of one event is calculated as the number of favorable outcomes divided by the total number of outcomes. In this case, the total number of outcomes is 6 (since a die has 6 faces), and the number of favorable outcomes is 2 (the numbers 5 and 6). So, the probability of getting a number greater than 4 in one toss is 2/6 = 1/3.

Step 2: Calculate the probability of two independent events The events are independent, meaning the outcome of the first event does not affect the outcome of the second event. The probability of two independent events both occurring is the product of their individual probabilities. So, the probability of getting a number greater than 4 on both tosses is (1/3) * (1/3) = 1/9.

So, the probability of having a number greater than 4 on each toss is 1/9.