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 and the specifics of a die.

1. A standard die has 6 faces, numbered from 1 to 6.
2. When the problem says "a number greater than 4", it means the outcomes we are interested in are 5 and 6. There are 2 such outcomes.
3. The total number of outcomes when a die is tossed is 6 (since a die has 6 faces).

The probability of an event is calculated as the number of favorable outcomes divided by the total number of outcomes.

So, the probability of getting a number greater than 4 in a single toss is 2/6 = 1/3.

Since the die is tossed twice, and these are independent events (meaning the outcome of the first toss does not affect the outcome of the second toss), we multiply the probabilities of the individual events to get the overall probability.

So, the probability of getting a number greater than 4 on both tosses is (1/3) * (1/3) = 1/9.