well, there are 216 possibilities, 1/6 x 1/6 x 1/6 = 1/216 so 216 possibilities. and there are 4 ways to get two or more fives, 155, 515, 551, and 555. so 4 out of 216 or 4/216, or 1/54. so there you go.
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I disagree. There are 16 ways out of 216 possibilities, so the probability is 16/216 or 2/27. The ways can represent by x55, 5x5, 55x and 555, where x can be any of five numbers not equal to five.
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It is 1 - 0.3120 = 0.6880, approx.
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.
if a die is rolled.What is the probibil ty that an even number divisible by 3 appears
When a tetrahedral die is rolled, it will rest with three faces upwards. If the die is numberd from 1 to 4. therefore the sum of the upward facing numbers on 1 die is at least 6 and so for two dice, the minimum is 12. That being the case, the probability is 0.
well, it will have 6 times of the greater chance.
It is 1 - 0.3120 = 0.6880, approx.
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.
The chance of any 1 number appearing in a die roll 3 times in a row is: 1/6 * 1/6 * 1/6 which is 1/216 = .00463
It is 0.1962
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]
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
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%