Math and Arithmetic

Statistics

Probability

Top Answer

If it is a regular dice then the probability is 3/6 that is 1/2

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0Because 3/6 of the sides on a number cube have even numbers, the probability of rolling even on one number cube is 1/2(equivalent of 3/6). But since you're rolling twice, you multiply the probability of one by itself (therefore rolling 2 number cubes). So: 1/2x1/2=1/4 The probability of rolling an even number when a number cube is rolled twice is 1/4, 25%, or 1 out of 4.

The probability is 1/6.

That depends on whether you roll it twice or not...

The probability of rolling 7 once with two dice is 1 in 6, o 0.1667. The probability, then, of doing that twice in a row is 1 in 36, or 0.02778.

Since you need a specific number - the number 6 - twice, the probability is (1/6)2, which is equal to 1/36.Since you need a specific number - the number 6 - twice, the probability is (1/6)2, which is equal to 1/36.Since you need a specific number - the number 6 - twice, the probability is (1/6)2, which is equal to 1/36.Since you need a specific number - the number 6 - twice, the probability is (1/6)2, which is equal to 1/36.

In 2 rolls, it's 1/6 = 162/3% if you don't care what the number is.If you name the number you want before rolling, it's 1/36 = 27/9%.

The probability of rolling a 4 in a die is 1 in 6, or about 0.1667. The probability, then, of rolling a 4 in at least one of two dice rolls is twice that, or 2 in 6, or 0.3333. The probability of rolling a sum of 4 in two dice is 3 in 36, or 1 in 18, or about 0.05556.

When a number cube is rolled twice, there are 36 possible outcomes. (1,1),(1,2),....(6,6). (3,3) occurs only once. Therefore, the probability of rolling a 3 both times is 1/36.

Let A be the event of rolling a 4. P(A) = 1/6 P(A)P(A)=(1/6)(1/6)=1/36 Therefore, the probability of rolling a 4 twice with two rolls of a number cube is 1/36.

The probability of getting an even number on both dice is (2/6) * (2/6) = (4/36) = 1/9. The probability when rolling two dice of getting one even number but not a "2" is 10/36, which is 5/18.

If you roll a die often enough, it is a certainty. If you roll a fair die just twice, the probability is 1/36.

The chance of rolling a 6 twice in a row, on a six-sided die, is 1 in 36 or 2.78%. The number of possible different results for rolling a six-sided die twice is 6 squared (6 times 6), or 36, therefore the probability of getting any one of the possible results is 1 out of 36.

Rolling it once there is a 1 out of 6 chance - because there is 6 numbers on a dice and one of them is three. so rolling it twice would be a 2 out of 12 chance / 0.16 / 16% chance

There are 6X6 or 36 combinations; and you can get a 1 & 6 twice; so the answer is 2/36 or 1/18.

The probability of rolling a sum of 8 on one roll of a pair of dice is 5/36.The probability of not rolling a sum of 8 on one roll of a pair of dice is 31/36.The probability of rolling a sum of 8 twice on two rolls of a pair of dice is(5/36)(5/36) = (5/36)2 .The probability of rolling first a sum of 8 and then rolling a sum that is not 8 on thesecond roll is (5/36)(31/36).The probability of rolling a sum that is not 8 on the first roll and rolling a sum of 8in the second roll is (31/36)(5/36).So The probability of rolling a sum of 8 at least one of two rolls of a pair of dice is(5/36)2 + (5/36)(31/38) + (31/36)(5/36) = 0.258487654... ≈ 25.8%.

I'm going to assume you mean rolling the same number twice in a row in 25 rolls. The first won't cause a double, so you just need to consider the odds of rolling the same number as the last for the last 24 rolls. The easier approach is to realize that the probability of rolling at least one double is 1 minus the probability of rolling no doubles. One roll has this probability of not rolling the same as the last: P(different number from last) = 5/6 Since they are independent events: P(no doubles in 25 rolls) = (5/6)24 Now the final probability, of at least one double, is 1 - (5/6)24

On a normal 6 sided dice the probability of rolling any number is 1/6.When we want a 3 AND then a 4 we multiply the probabilities together.So P(3 then 4) = 1/6 x 1/6 = 1/36

If you mean rolling at least one three, the 1/3. If you mean exactly one three, then 5/18. and If you mean a sum of 3 for the two throws, the answer is 1/18

Assuming that a dice roll is purely random, there is a 1 in 6 probability of landing on any number. Since the second roll depends on the probability of the first, they factor together: First Roll: 1/6 Second Roll: (1/6)(1/6) = 1/36 If you were rolling both dice at once, however, the math would be completely different.

The probability is 0.0198. The probability of that happening twice is 0.000393 Is this the odds of it happening twice with the same number?

First calculate the probability of NOT getting a single 5. This probability is 5/6 x 5/6 = 25/36. Therefore, the probability of getting at least one 5 is the complement thereof, that is, 1 - 25/36 = 11/36.

2/36 what you rolled the first time has nothing to do with the next roll. It's an independent event.

Since there is a one in six chance of rolling a four on any given roll the probability is 1/6 or 0.16666... or 16.7%. Since each roll is independent the probabilities can be multiplied. For example to roll a four twice in a row is 1/6 x 1/6 or (1/6)^2. Note how quickly the probability drops. So, for a six sided dice the formula for rolling any number N times in row is (1/6)^N.

The chance is one in 216 (6^-3).The probability of rolling a five once is 1/6. Rolling a five again, on the same die or another, will still have a 1/6 chance. Therefore, the probability of the event occurring twice is 1/36 (1/6^2). Three times has a probability of 1/216 (1/6^3), and so on. It does not matter what die is used, as long as it has six sides.The probability p of rolling a number x times consecutively on an s sided die isp=s^-x

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