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.
The probability of getting an even number on at least one of the 3 rolls is 7/8.
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.
Possible outcomes of one roll = 6.Probability of an even number on one roll = 3/6 = 0.5 .Probability of an even number on the second roll = 0.5 .Probability of an even number on the third roll = 0.5 .Probability of an even number on all three rolls = (0.5 x 0.5 x 0.5) = 0.125 = 1/8The probability of at least one odd number is the probability of not gettingan even number on all 3 rolls. That's (1 - 1/8) = 7/8 or 0.875 or 87.5% .
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.
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
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 probability of getting an even number on at least one of the 3 rolls is 7/8.
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%