if a die is tossed four times what is the probability of getting anything but a five on every toss

5/6 on each toss, right?

5/6 * 5/6 * 5/6 * 5/6 = answer

= (5/6)^4 = .482

To find the probability of getting anything but a five on each toss of a die, we first need to understand the total number of possible outcomes and the favorable outcomes.

Total number of possible outcomes when a die is tossed = 6 (since a standard die has six faces labeled 1, 2, 3, 4, 5, and 6)

Next, we need to determine the number of favorable outcomes, which is the number of outcomes where anything but a five is obtained on each toss. In this case, the favorable outcomes include obtaining any number except 5 on each of the four tosses.

Number of favorable outcomes = Number of possible outcomes for each toss (which is 5) raised to the power of the number of tosses (which is 4)

Number of favorable outcomes = 5^4 = 625

Now, we can calculate the probability:

Probability = Number of favorable outcomes / Number of total outcomes

Probability = 625 / 6^4

Probability = 625 / 1296 ≈ 0.4815 (rounded to four decimal places)

Therefore, the probability of getting anything but a five on every toss of a die is approximately 0.4815.