The answer depends on how many rolls.
If there were n rolls, then the probability is n*(1/6)*(5/6)n-1/[1 - (5/6)n]
It is 0.722... recurring.
The probability is 0.6187, approx.
It is 0.9459
you toss 3 coins what is the probability that you get exactly 2 heads given that you get at least one head?
It is 0.1962
It is 0.722... recurring.
The probability is 0.6187, approx.
It is 0.9459
The probability that 14 is rolled at least once is 1 - 5.5*10-32 which, for all intents and purposes, can be treated as 1.
When rolling one die, the probability of getting a 4 is 1 in 6, or 0.1667. If two dice are rolled, you get two unrelated chances of rolling at least one 4, so the probability is 2 in 6, or 0.3333.
you toss 3 coins what is the probability that you get exactly 2 heads given that you get at least one head?
It is 0.1962
The probability is 1 and you do not need Matlab to get that answer - only a little bit of thought.
3/16 Michigan Virtual says it is 11/12
If you mean rolling at least one three, the 1/3. If you mean exactly one three, then 5/18. and If you mean a sum of 3 for the two throws, the answer is 1/18
It is 1 - 0.3120 = 0.6880, approx.
Probability(at least 1 six) = 1 - probability(no sixes) = 1 - 5/6 x 5/6 x 5/6 = 1 - 125/216 = 91/216 ≈ 0.421 = 42.1%