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Q: If two twelve side fair dice are rolled what it the probability that the total number of spots shown is equal to 5?

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Since all numbers on a die are 6 or below, the probability is 6/6, 1, or 100%.

If two dice are rolled, the probability that the sum of observed values is equal to 13 is zero. The sum of two dice can only lie between 2 and 12, inclusive.

It is: 1/6 Since a die is cubic in shape and has 6 sides, and there is an equal probability of getting any one of them facing up (assuming the die is not loaded), and there is only one number on a die greater than 5, that being 6, the probability of rolling such a number is therefore 1 in 6, or .1666...

Number of possible total pairs = 36 Number of pairs which give five as a sum = {1,4},{2,3},{3,2},{4,1} So, P(Sum=5) = 4/36 = 1/9

It is 1.

Related questions

The probability to get a 12, with two dice, is 1/36.

If two fair dice were rolled, there would be 36 outcomes. (1,1),(1,2),....,(6,6) The maximum sum would be 12. Therefore, the probability that the total number of spots shown is equal to 15 is zero.

Since all numbers on a die are 6 or below, the probability is 6/6, 1, or 100%.

If two dice are rolled, the probability that the sum of observed values is equal to 13 is zero. The sum of two dice can only lie between 2 and 12, inclusive.

The number 12.

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.

It is: 1/6 Since a die is cubic in shape and has 6 sides, and there is an equal probability of getting any one of them facing up (assuming the die is not loaded), and there is only one number on a die greater than 5, that being 6, the probability of rolling such a number is therefore 1 in 6, or .1666...

6 & 4 And 5 & 5 are the only two possibilities, so 4 chances in 36 or 1 in nine.

Larger or equal to 0 but less than or equal to 1

Number of possible total pairs = 36 Number of pairs which give five as a sum = {1,4},{2,3},{3,2},{4,1} So, P(Sum=5) = 4/36 = 1/9

It is 1/3

! in 4, as the four suits have an equal number of cards.

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