125/3888 or 3.22%.
A shortcut can be taken instead of the traditional tree-method or combination counting:
Assume that there are 5 "spots" in which the 3 sixes can be placed. Using the nCr probability formula, we find that there are 10 different arrangements of 3 sixes in 5 spots. (5!/((5-3)!*3!)). This simplifies to (5*4)/2, which is 10.
A visual of this:
6-6-6-X-X
6-6-X-6-X
6-6-X-X-6
6-X-6-6-X
6-X-6-X-6
6-X-X-6-6
X-6-6-6-X
X-6-6-X-6
X-6-X-6-6
X-X-6-6-6
Next, all we need to do is figure out how many different combinations of each of the above combinations there are. Since there are only two variable numbers (denoted X) in each combination, and each can only be an integer from 1-5, we can determine that for each combination above there are 25 subsequent combinations (5^2).
Finally, (10*25)/(6^5) = 125/3888.
At least 2 sixes mean 2,3,4, or 5 sixes. This is the sum of the probability of 2 sixes + probability of 3 sixes etc. It is easier to look at the probability of 0 or 1 six and subtract that the sum of those two from 1. P( one six)=1/6 P(number other than six)=5/6 P( one six in 5 tosses)=(1/6)(5/6)^4=625/7776 P (no six in 5 tosses) the probability of not getting a 6 of the first toss is 5/6 Now if we do not get one on the first toss, we have 5/6 chance of not getting it again on the second, and the third etc. (5/6)^5=3125/7776 If we add these we have 3750/7756 which is the odds of exactly 0 or 1 six. 1-(3750/77756)=2003/3878 .516503 is the probability of at least 2 sixes in 5 tosses.
If you toss the die often enough then the probability of getting the sequence 2-2-1 is 1: a certainty. The probability of getting the result in the first three tosses is 1/216.
It is a certainty. If the die is rolled often enough, the probability that two consecutive rolls show a six is 1.
The probability of flipping a quarter and getting heads is 1 in 2. the probability of rolling a die and getting 6 is 1 in 6.
It depends on what size die you use, what its labels are and how many rolls you make. For example using a standard six-sided die and one roll, the probability of no sixes is 5/6 or ~0.83; the probability of no sixes with 25 rolls is less than 0.01 or 1%. If you used a standard d3 (three-sided die) then the probability will always be 1 or 100%, since rolling a six is impossible; but if every side has '6' on it the probability is 0, since every roll must be a 6.
The probability is 0.0322
At least 2 sixes mean 2,3,4, or 5 sixes. This is the sum of the probability of 2 sixes + probability of 3 sixes etc. It is easier to look at the probability of 0 or 1 six and subtract that the sum of those two from 1. P( one six)=1/6 P(number other than six)=5/6 P( one six in 5 tosses)=(1/6)(5/6)^4=625/7776 P (no six in 5 tosses) the probability of not getting a 6 of the first toss is 5/6 Now if we do not get one on the first toss, we have 5/6 chance of not getting it again on the second, and the third etc. (5/6)^5=3125/7776 If we add these we have 3750/7756 which is the odds of exactly 0 or 1 six. 1-(3750/77756)=2003/3878 .516503 is the probability of at least 2 sixes in 5 tosses.
If you toss the die often enough then the probability of getting the sequence 2-2-1 is 1: a certainty. The probability of getting the result in the first three tosses is 1/216.
2 out of 6
The probability of tossing a die and getting three 6's in a row is (1/6)3, or about 0.004630.
Im not sure what u are asking the probability of getting 2 sixes on both dice is1/36 the probability of getting a six on either dice is 1/6 and the probability of getting one six on both of the dice is 5/36 i believe hope i answered ur question
.027777778 or 1 in 36.
It is a certainty. If the die is rolled often enough, the probability that two consecutive rolls show a six is 1.
It is 0.1962
there are 36 possible combinations in two single die tosses. The odds of any one combination is then 1:36
The probability of getting a six on a six sided die and then getting a tails is zero. There is no tails on a die.
I'm assuming you're looking for the probability that you roll either a one or six at least once. So the problem can be rewritten as: 1 - probability of rolling 60 times and never getting ones or sixes = 1 - (2/3)^60