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Two dice are rolled The probability that their sum is at least 9 equals?
Outcomes from rolling two dice: Total Probability 2 1/36 3 2/36 4 3/36 5 4/36 6 5/36 7 6/36 8 5/36 9 4/36 10 3/36 11 2/36 12 1/36 So, the probability of rolling 9 or greater is 4/36+3/36+2/36+1/36 = 10/36 ~= 0.278 = 27.8%
When two dice are rolled what is the probability of getting a sum of 5 or 6?
When two dice are rolled, the probability of getting: 2 is 1/36 3 is 2/36 4 is 3/36 5 is 4/36 6 is 5/36 7 is 6/36 8 is 5/36 9 is 4/36 10 is 3/36 11 is 2/36 12 is 1/36
The outcomes for the sum of two dice can be described as a discrete uniform distribution?
No, it resembles a normal distribution, but discrete. sum --- Probability 2-----------1/36 3-----------2/36 4-----------3/36 5-----------4/26 6-----------5/36 7-----------6/36 8-----------5/36 9-----------4/36 10----------3/36 11----------2/36 12----------1/36
What is the probability that the sum of the numbers from two rolled dice result in a perfect square or even number?
Part1: Finding probability of getting sum as a perfect square. Maximum sum of both the dice is (6+6) equal to 12. Up to 12, the perfect squares are: 1, 4 and 9. Getting a sum of 1 from two dice is not possible. So, we are left with 4 and 9. To get 4, the combination can be: (2,2) or (1,3) or (3,1). This means, to get the sum as 4, the probability is [3/36]. To get 9, the combination can be: (3,6) or (6,3) or (5,4) or (4,5). This means, to get the sum as 9, the probability is [4/36]. Therefore,the total probability of getting the sum as a perfect square is: [(3/36)+(4/36)]=[7/36]. Part2: Finding the probability of getting sum as an even number. The possible even numbers can be 2, 4, 6, 8, 10 and 12. But, as 4 is already considered in part1, it should be ignored in this case. The probability of getting sum as 2 is: [1/36] The probability of getting sum as 6 is: [5/36] The probability of getting sum as 8 is: [5/36] The probability of getting sum as 10 is: [3/36] By adding all the above, the probability of getting sum as an even number (ignoring 4) is: [(1/36)+(5/36)+(5/36)+(3/36)]=[14/36]. From part 1 and part 2, we get the total probability as [(7/36)+(14/36)]=[7/12]=0.583333.
What is the probability of rolling a sum that is a multiple of 4 or a set of doubles with two number cubes?
Total possibilities with one cube = 6.Total possibilities with two cubes = 6 x 6 = 36.Possible multiples of 4 = 8 possibilities:1 - 32 - 22 - 63 - 13 - 54 - 45 - 36 - 2Doubles not already included = 3 possibilities:3 - 35 - 56 - 6Total possible successes = 8 + 3 = 11 .Probability of success = 11/36 = 30.6%(rounded)
