3 fair dice are tossed. Find the probability of dice showing at most one 6
To find the probability of dice showing at most one 6, we need to calculate the number of favorable outcomes and divide it by the total number of possible outcomes.
Let's consider the possible outcomes when tossing three fair dice.
Each die has six possible outcomes: 1, 2, 3, 4, 5, or 6. Since we have three dice, the total number of possible outcomes is 6 * 6 * 6 = 216.
Now let's determine the number of favorable outcomes, which are the outcomes where dice show at most one 6.
Case 1: No 6 is rolled
In this case, each die can show any number from 1 to 5. So, the number of favorable outcomes is 5 * 5 * 5 = 125.
Case 2: One 6 is rolled
There are three ways to choose which die shows the 6, and the remaining two dice can show numbers from 1 to 5. So, the number of favorable outcomes is 3 * 5 * 5 = 75.
Therefore, the total number of favorable outcomes is 125 + 75 = 200.
Finally, we divide the number of favorable outcomes by the total number of possible outcomes to calculate the probability:
Probability = (Number of Favorable Outcomes) / (Total Number of Possible Outcomes)
= 200 / 216
≈ 0.9259
So, the probability of dice showing at most one 6 is approximately 0.9259, or 92.59%.