Chuck a luck is a game of chance in which 3 dice in a cage are tumbled. Bets can be placed on the values that the 3 dice will show. Find the probability that none of the dice will show a 5
Chuck a luck is a game where each of the 3 dice has 6 possible outcomes (1, 2, 3, 4, 5, or 6) and there are a total of 6^3 = 216 possible outcomes when the dice are tumbled.
To find the probability that none of the dice will show a 5, we need to determine the number of outcomes in which none of the dice show a 5.
Since each die has 5 possible outcomes that are not 5 (1, 2, 3, 4, or 6), there are 5^3 = 125 outcomes in which none of the dice show a 5.
Therefore, the probability that none of the dice will show a 5 is 125/216, which can be simplified to approximately 0.579 or 57.9%.
To find the probability that none of the dice will show a 5, we need to calculate the number of favorable outcomes and divide it by the total number of possible outcomes.
First, let's determine the total number of possible outcomes. Since we have 3 dice and each die has 6 possible outcomes (numbers 1 to 6), the total number of possible outcomes is 6 x 6 x 6 = 216.
Now, let's calculate the number of favorable outcomes, i.e., the outcomes where none of the dice shows a 5. There are 5 possible outcomes for each die (numbers 1, 2, 3, 4, and 6), so the number of favorable outcomes is 5 x 5 x 5 = 125.
Therefore, the probability that none of the dice will show a 5 is 125/216, which simplifies to approximately 0.5787, or 57.87%.