When a die is rolled once, the probability of a 4 showing up is 1/6.
Apply the binomial probability for finding the probability of exactly three fours out of 12 throws of a die.
n=12 (number of throws)
p=1/6 (probability of a four in a single throw)
x = 3 (number of times out of 12 , a four showing up)
P(x=3) = 12C3 (1/6)^3 (5/6)^(12-3)
= 12C3 (1/6)^3 (5/6)^9 = 0.197443
3/18 I think
It is approx 0.1974
Assuming the die is a standard die with a different number from the set {1, 2, 3, 4, 5, 6} on each side, then: Probability_of_success = number_of_ways_of_success/total_number_of_outcomes There is only 1 way to roll a 4 and there are 6 possible outcomes, therefore the probability of rolling a 4 is 1/6 The die has no knowledge of previous rolls; each roll is independent, thus: The probability of three fours is a row is the probability of a 4 times the probability of a 4 times the probability of a 4, which is: Probability = 1/6 × 1/6 × 1/6 = 1/216
The probability of 3 specific dice rolls is the probability that each one will happen multiplied together. For instance, the probability of rolling 2 then 6 then 4 is the probability of all of these multiplied together: The probability of rolling 2 is 1/6. The probability of rolling 6 is 1/6. The probability of rolling 4 is 1/6. Multiply these together and we get the total probability as 1/216
Anywhere from 0 to 1; it depends on the shape and what numbers are written on the faces.
3/18 I think
The probability of getting three fives in the first three rolls and non-fives in the next three rolls is; P(5,5,5,N5,N5,N5) = 1/6 x 1/6 x 1/6 x 5/6 x 5/6 x 5/6 = 0.002679... The number of different order in which the fives can come out is given by; 6C3 = 6!/[3!∙(6-3)!] = 20 So the probability that in 6 rolls of a fair die exactly three fives (in any order) will come out is; P(three fives any order) = (20)∙(1/6)3∙(5/6)3 = 0.05358... ~ 5.4%
With a fair die, it is 1/216 in three rolls, but the probability increases to 1 (a certainty) as the number of rolls is increased.
It is approx 0.1974
Assuming the die is a standard die with a different number from the set {1, 2, 3, 4, 5, 6} on each side, then: Probability_of_success = number_of_ways_of_success/total_number_of_outcomes There is only 1 way to roll a 4 and there are 6 possible outcomes, therefore the probability of rolling a 4 is 1/6 The die has no knowledge of previous rolls; each roll is independent, thus: The probability of three fours is a row is the probability of a 4 times the probability of a 4 times the probability of a 4, which is: Probability = 1/6 × 1/6 × 1/6 = 1/216
The probability of 3 specific dice rolls is the probability that each one will happen multiplied together. For instance, the probability of rolling 2 then 6 then 4 is the probability of all of these multiplied together: The probability of rolling 2 is 1/6. The probability of rolling 6 is 1/6. The probability of rolling 4 is 1/6. Multiply these together and we get the total probability as 1/216
The probability is 90/216 = 5/12
Anywhere from 0 to 1; it depends on the shape and what numbers are written on the faces.
33%
Rolling a sum of 15 on three rolls of a die, when the first roll is a 4, is the same as rolling a sum of 11 on the second and third roll. The probability of rolling 11 on two dice is 3 in 36, or 1 in 12.
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.
It is 0.1042