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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 the

second 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 8

in 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%.

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Q: What is the probability of rolling a sum of 8 on at least one of two rolls of a pair of number cubes?

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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 assuming you're looking for the probability that you roll either a one or six at least once. So the problem can be rewritten as: 1 - probability of rolling 60 times and never getting ones or sixes = 1 - (2/3)^60

The least error prone way of doing this is to draw a 6x6 square of all the possiblities when rolling a pair of dice, where one is red and one blue, and work out the proportion of the combinations that sum to 7.

The probability of rolling a number divisible by three on a fair die (3 and 6) is 2 in 6, or 1 in 3, or about 0.3333. Since there are 36 permutations of rolling two dice, the probability, then, of rolling (at least one) number divisible by three on two rolls of a die is 19 in 36, or about 0.5278. Simply count the number of desired results, and divide by the number of possible results, as I have done below.111213 x141516 x212223 x242526 x31 x32 x33 x34 x35 x36 x414243 x444546 x515253 x545556 x61 x62 x63 x64 x65 x66 xIf, on the other hand, you are looking at the probability of rolling a sum divisible by three, consider ...11 212 3 x13 414 515 6 x16 721 3 x22 423 524 6 x25 726 831 432 533 6 x34 735 836 9 x41 542 6 x43 744 845 9 x46 1051 6 x52 753 854 9 x55 1056 1161 762 863 9 x64 1065 1166 12 x... which is 12 in 36, or 1 in 3, or about 0.3333.

The probability of rolling a 6 on each roll of an unbiased cuboid die is 1/6 If you mean at least one of the rolls shows a 6 then it is the same as 1 - pr(no roll shows a 6) = 1 - (5/6)⁶⁰ ≈ 1 - 0.0000177 = 0.9999823 If you mean that exactly one 6 is rolled then: Pr(exactly one 6) = 60 × 1/6 × (5/6)⁵⁹ ≈ 0.0002130

Related questions

It is 5/18.

one fourth

With one roll of three dice, the probability is 7/8.

00

The probability of rolling at least one 2 when rolling a die 12 times is about 0.8878. Simply raise the probability of not rolling a 2 (5 in 6, or about 0.8333) to the 12th power, getting about 0.1122, and subtract from 1.

The probability of rolling at least one '1' with six dice is 66.5% [1-(5/6)^6]*100%

It is 0.9459

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

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.

11/12

Depends on what you are rolling - a cubic die or a more exotic shape.

The probability of rolling a six is 1 out of 6, or 1/6. Now, perhaps your question is: If I roll a die 180 times, what is the probability of rolling a six at least once. This is the same as rolling a die 180 times and never once rolling a six. The probability is (5/6)180 which is 5.59 x 10-15.

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