1/6= 2 because there is only one 2.
Therefore the theoretical probability of not rolling a two is the same as everything but two so 5/6.
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Assuming you are talking about ordinary dice the odds of rolling 2 are 1 chance in 36
The probability of rolling an odd number on a standard die is 3 in 6, or 1 in 2, or 0.5.
The probability of rolling at least one 2 in fifty rolls of a standard die is 1 - (5/6) 50, or about 0.99989012. This calculation starts by looking at the probability of not rolling a 2, which is 5/6. To repeat that 50 times in a row, you simply raise that to the 50th power, getting 0.000109885. Then you subtract the result from 1 to get the probability of not succeeding in not rolling a 2 in fifty tries. Expressed in normal "odds" notation, this is about (100000 - 11) in 100000, or about 99989 in 100000.
The factors of 10 are the numbers that divide 10 evenly: 1, 2, 5 and 10. To answer your question, you have to figure out what the probability of rolling one of these numbers is on a number cube.
The probability of rolling at least one 2 when rolling a die 12 times is about 0.8878. Simply raise the probability of not rolling a 2 (5 in 6, or about 0.8333) to the 12th power, getting about 0.1122, and subtract from 1.
No probability - theoretical or not - can be 100. Therefore no examples are possible.No probability - theoretical or not - can be 100. Therefore no examples are possible.No probability - theoretical or not - can be 100. Therefore no examples are possible.No probability - theoretical or not - can be 100. Therefore no examples are possible.