If you mean 1: head 2:tail 3:head the probability is 1/2 x 1/2 x 1/2 = 1/8.
If you mean only that you want the heads and tails to alternate, that is, you mean the above or tail, head , tail, the probability is 1/8 + 1/8 = 1/4.
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.
The opposite of getting at most two heads is getting three heads. The probability of getting three heads is (1/2)^2, which is 1/8. The probability of getting at most two heads is then 1 - 1/8 which is 7/8.
In the long run, it is a certainty. In the first three throws it is 1/216.
The probability is 1/4
The probability of getting an even sum on two dice is 18 in 36 or 1 in 2 or 0.5. The probability of doing that three times in a row is 0.53 or 0.125.
Probability = (1/6)3 = 1/216 =0.00463
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.
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.
Since the probability of getting tails is 50% or 0.5, the probability of three tails would be 0.5*0.5*0.5=0.125 or 12.5 %
The probability is 0.0322
The opposite of getting at most two heads is getting three heads. The probability of getting three heads is (1/2)^2, which is 1/8. The probability of getting at most two heads is then 1 - 1/8 which is 7/8.
In the long run, it is a certainty. In the first three throws it is 1/216.
The probability is 1/4
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
The probability of getting an even sum on two dice is 18 in 36 or 1 in 2 or 0.5. The probability of doing that three times in a row is 0.53 or 0.125.
The probability is 3/8.The probability is 3/8.The probability is 3/8.The probability is 3/8.
1/6^3