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What is the probability of throwing a sum of 15 on three rolls of a fair die if the first roll was a 4?
Wiki User
∙ 13y ago
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.
Wiki User
∙ 13y ago
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What is the probability that 6 rolls of a fair die will show exactly three fives?
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%
What is the probability that none of the rolls is a six?
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.
What is the probability of drawing three aces in a row without replacement?
The probability of drawing aces on the first three draws is approx 0.0001810
A dice is rolled three times the probability of getting at least one odd number is?
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% .
What is the probability of throwing dice in the order of 1 2 and 6?
1/120there are three spots and six numbers on a diceso 6 numbers on the first spot and 5 numbers on the second spot and 4 in the third spot just multiply them
What is the probability of rollling a 5 3times ina row?
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.
What is the probability of throwing three twos with two dice?
The probability is zero since at most there could be 2 two's.
What is the probability of throwing a 1 a 2 and then a 6 in exactly that order?
If a fair die is thrown often enough, the probability is 1.For the first three throws of a fair die, the probability is 1/216.If a fair die is thrown often enough, the probability is 1.For the first three throws of a fair die, the probability is 1/216.If a fair die is thrown often enough, the probability is 1.For the first three throws of a fair die, the probability is 1/216.If a fair die is thrown often enough, the probability is 1.For the first three throws of a fair die, the probability is 1/216.
What is the probability that 12 rolls of a fair die will show three ones?
It is approx 0.1974
What is the probability of rolling exactly one 4 in three rolls of a dice?
3/18 I think
How do you get the probability of three dice rolls?
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
How do you figure out what the probability of rolling a three and then a four using a standard dice?
If you keep rolling the die, the probability is 1. If you require a 3 and a 4 in the first two rolls, the answer is (1/6)*(1/6) = 1/36
What is the probability that 6 rolls of a fair die will show exactly three fives?
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%
You roll six sided die three timeswhat is the probability that all three rolls will be a 1?
The probability that each roll will be a 1, is 1/6 (a sixth) because there is one outcome of interest (getting a 1) and 6 possible outcomes (6 numbers on the die).Probability rules mean that if you want the probability of getting outcome A and getting outcome B then the total probability is P(A) x P(B) where P(A) means the probability of getting outcome A).In short if you want P(A and B) then this is P(A) x P(B)Applied to this example if you want the probability of getting a 1 on each throw of the die (i.e. on all 3 throws) then the probability is given by:P(1 on all three rolls) = P(1 on first roll) x P(1 on second role) x P(1 on third role)P(1 on all three rolls) = 1/6 x 1/6 x 1/6P(1 on all three rolls) = 1 / 216
What is the probability of getting three fives in three rolls of a single die?
Anywhere from 0 to 1; it depends on the shape and what numbers are written on the faces.
What is the probability that 13 rolls of a fair die will show three threes?
Not an easy question. Here's a long but logical answer. The chance of not getting a 3 on any throw is 5/6, so the chance of not getting any threes in the first 10 throws is (5/6)^10. Now, having reached that point, the chance of throwing three threes in the last three throws is 1/6 x 1/6 x1/6 or 1/216. So the probability of throwing three threes in any chosen three throws is (5/6)^10 x 1/216. This should seem logical because the probability of throwing three threes in the first three throws followed by ten failures to throw threes, should be the same as just getting threes in the last three throws. The probability is the same for getting threes in only the second, fourth and sixth throw, or any other combination of three throws. Now, there are a set number of combinations of 13 things taken in groups of three. This is given by the formula for C(13,3) = (13x12x11x10x9x8x7x6x5x4)/((10x9x8x7x6x5x4x3x2x1)x(3x2x1)). That cancels down to (13x12x11)/(3x2x1), and equals 286. So, there are this many ways you can pick out three throws which you want to be the threes, and the probability of them actually being threes is the first calculation we did. The answer to the question is 286 x (5/6)^10 x 1/216 which is 0.2138453, about.
What is the probability that none of the rolls is a six?
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.
