If it is a regular dice then the probability is 3/6 that is 1/2
It is 1/36.
That depends on whether you roll it twice or not...
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
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
Because 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.
If it is a regular dice then the probability is 3/6 that is 1/2
The probability is 1/6.
50 percent
It is 1/36.
That depends on whether you roll it twice or not...
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.
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
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
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.
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.
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%.