There are 3 odd numbers since there are a total of 6 consecutive numbers on a dice. For example, the numbers 1 2 3 4 5 6 are on a dice, so the odd numbers are 1, 3, and 5.
163759438475047695685695034395784593850835943859458474836937657 i have just tried it
2, 3, 5
11 outcomes if the dice are indistinguishable, 36 otherwise.
If any numbers come up more often than any others over the long run, then the dice are loaded and the game is not honest. * * * * * That is true for the roll of a die, but not dice. With 2 dice, the probabilities of the numbers are as follows 2 or 12 : 1/36 each 3 or 11 : 2/36 or 1/18 each 4 or 10 : 3/36 or 1/12 each 5 or 9 : 4/36 or 1/9 each 6 or 8 : 5/36 each 7 : 6/36 or 1/6 So 7 is the most rolled number in the long run when 2 dice are rolled.
There are 21 combinations.
There are 3 odd numbers since there are a total of 6 consecutive numbers on a dice. For example, the numbers 1 2 3 4 5 6 are on a dice, so the odd numbers are 1, 3, and 5.
Three of them and they are 2, 4 and 6
there are (2, 3, 5) only 3 prime numbers on a single dice. 1 is not considered a prime number.
2 each of the numbers 1 thru 6. ---------------------------------- This can be answered in many ways: --- There are twelve faces on two dice, each with a number --- The are six different numbers on the dice faces --- There are 36 possible combinations of two dice --- --- They can total between 2 and 11 --- --- There are 21 different possible pairs (all but six of which can appear two ways)
Eight numbers.
The question is underspecified since the answer depends on the numbers on the dice. If all the numbers on both the dice are the same, there is clearly only one outcome. If the dice have 4 different numbers, then there can be 16 different outcomes. If the numbers on each die are 1,2,3 and 4 (or any four numbers in arithmetic sequence) there will be 7 outcomes.
3
6
3
Two.Two.Two.Two.
I'm assuming your question is the same as this: "If 2 dice are rolled, what is the probability of not getting 1 on either die?" To answer this question, we need to look at what IS possible. If I'm 2 rolling normal, fair dice, then I have equal probability of getting each of the numbers 1-6 on either die. If I'm trying to NOT get 1, then I want to get any of the numbers 2-6 on both dice. This gives me 10 desired outcomes (5 numbers * 2 dice) out of 12 possible outcomes (6 numbers * 2 dice), so the probability is 10/12, which simplifies to 5/6.