Assuming that the die is fair rather than far, the probability is (5/6)9(1/6) = 1,953,125/60,466,176 = 0.0323 approx.
To find the experimental probability of rolling a 6, you first need to determine the number of times a 6 was rolled during the experiment. Then, divide that number by the total number of rolls recorded in the table. The resulting fraction represents the experimental probability of rolling a 6. For example, if a 6 was rolled 5 times out of 30 total rolls, the experimental probability would be 5/30, which simplifies to 1/6.
You have rolled doubles so the full set of possible outcomes is [1,1], [2,2], [3,3], [4,4],[5,5] and [6,6] - that is six in all. Of these the first four are favourable outcomes (sum < 9) So the required probability is 4/6 or 2/3
2/36 what you rolled the first time has nothing to do with the next roll. It's an independent event.
The probability of rolling a 2 or a 4 on the second die is independent of the outcome of the first die. A standard die has six faces, with three even numbers (2, 4, and 6). Therefore, the probability of rolling a 2 or a 4 on the second die is 2 out of 6, or 1/3, regardless of whether an even number was rolled on the first die.
There are 36 possible combinations of unique dice rolls (where order doesn't matter, so sorted/deduped). 6 of these are potentially doubles. The total odds are 1/6 (~16.7%). This is because the probability that any of the two die matched any of the other two dice is really just the probability that a die is the same as another die, which is 1/6. Think about it this way, if you rolled two dice the probability that the first one is the same as the second one is 1/6. If you rolled a one, and then rolled a second die and it didn't come up as the first one, but then you rolled a third one you'll find that the probability that doubles occurred is the probability that the first one matches the third one, OR the second one matches the third one IIF the second one did not match the first one, but because they're independent events it's still just what's the probability that doubles occurred between the comparison rather than the set. This does not take into account triplets being considered doubles. If the potential for triplets is considered (the probability that any two die will have the same value), this becomes a very simple problem because it deals with the likelihood that something can't occur in any form rather than the likelihood that a specific subset will occur. It's simply the probability (after sorting) that the first one is like the second one, or the second one is like the third one. When you think about that, that's really just the probability that all numbers will be different. This means we only have to care about two. The probability that the second is different is 5/6, while the probability that the third is different than both of the first two is 4/6. That makes the probability that no two are alike 5/6 by 4/6. This means that the probability that AT LEAST two are alike is 1 - (5/6 * 4/6), which is 44.444..%. It's an extremely simplified version of the birthday problem.
For the first 5 to appear on the tenth roll, the first nine rolls must not be a 5 and the tenth must. Therefore: probability = (5/6)9 x 1/6 ≈ 0.0323 = 3.23 %
To find the experimental probability of rolling a 6, you first need to determine the number of times a 6 was rolled during the experiment. Then, divide that number by the total number of rolls recorded in the table. The resulting fraction represents the experimental probability of rolling a 6. For example, if a 6 was rolled 5 times out of 30 total rolls, the experimental probability would be 5/30, which simplifies to 1/6.
The first roll doesn't matter for probability, it just sets the number to be rolled by the other two. So: P(rolling the same number three times) = P(rolling a particular number)2 = (1/6)2 = 1/36
You have rolled doubles so the full set of possible outcomes is [1,1], [2,2], [3,3], [4,4],[5,5] and [6,6] - that is six in all. Of these the first four are favourable outcomes (sum < 9) So the required probability is 4/6 or 2/3
Three out of twelve
2/36 what you rolled the first time has nothing to do with the next roll. It's an independent event.
First of all it is probability second of all the answer is 1/6+1/6 which is 2/12 which simplified is 1/6
odds in favor is when the number you want is first and the number out of is last.
The probability of rolling a 2 or a 4 on the second die is independent of the outcome of the first die. A standard die has six faces, with three even numbers (2, 4, and 6). Therefore, the probability of rolling a 2 or a 4 on the second die is 2 out of 6, or 1/3, regardless of whether an even number was rolled on the first die.
There are 36 possible combinations of unique dice rolls (where order doesn't matter, so sorted/deduped). 6 of these are potentially doubles. The total odds are 1/6 (~16.7%). This is because the probability that any of the two die matched any of the other two dice is really just the probability that a die is the same as another die, which is 1/6. Think about it this way, if you rolled two dice the probability that the first one is the same as the second one is 1/6. If you rolled a one, and then rolled a second die and it didn't come up as the first one, but then you rolled a third one you'll find that the probability that doubles occurred is the probability that the first one matches the third one, OR the second one matches the third one IIF the second one did not match the first one, but because they're independent events it's still just what's the probability that doubles occurred between the comparison rather than the set. This does not take into account triplets being considered doubles. If the potential for triplets is considered (the probability that any two die will have the same value), this becomes a very simple problem because it deals with the likelihood that something can't occur in any form rather than the likelihood that a specific subset will occur. It's simply the probability (after sorting) that the first one is like the second one, or the second one is like the third one. When you think about that, that's really just the probability that all numbers will be different. This means we only have to care about two. The probability that the second is different is 5/6, while the probability that the third is different than both of the first two is 4/6. That makes the probability that no two are alike 5/6 by 4/6. This means that the probability that AT LEAST two are alike is 1 - (5/6 * 4/6), which is 44.444..%. It's an extremely simplified version of the birthday problem.
First calculate the probability of not rolling a six - since there are 5 possibilities for each die, this is (5/6) x (5/6). Then calculate the complement (1 minus the probability calculated).First calculate the probability of not rolling a six - since there are 5 possibilities for each die, this is (5/6) x (5/6). Then calculate the complement (1 minus the probability calculated).First calculate the probability of not rolling a six - since there are 5 possibilities for each die, this is (5/6) x (5/6). Then calculate the complement (1 minus the probability calculated).First calculate the probability of not rolling a six - since there are 5 possibilities for each die, this is (5/6) x (5/6). Then calculate the complement (1 minus the probability calculated).
Since there are 6 sides on every die that are equally likely to be rolled, the probability of rolling any given side once is exactly 1/6. The 2 events or the first and second dice roll are independent (the outcome of one does not influence the other) so to find the probability of both occurring you just multiply the probability of each event. Since each event has a 1/6 probability of occurring as stated before, The entire event has a probability of 1/6*1/6 or 1/36, which is approximately 2.78%.