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Q: What is the probability of getting a sum of 6 or 7 or 8?

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Probability that the sum is 6 = 5/36 Probability that the sum is 7 = 6/36

The probability that the sum of two dice is 7 is 6 in 36, or 1 in 6.Of all the combinations, this is the one with the highest probability.

There are 36 permutations of two dice. Of them, 6 have a sum of 7, (1+6, 2+5, 3+4, 4+3, 5+2, and 6+1) and 2 have a sum of 11 (5+6 and 6+5). The probability, then of getting a sum of 7 or 11 is (6 plus 2) in 36, or 8 in 36, or 2 in 9, or about 0.2222.

First we find the probability of getting a 7. Of the 36 outcomes possible 6 result in a sum of 7, in other words 1/6. The probability of getting an 11 is 2/36 or 1/18. The probability of getting one or the other is the sum of the two, 8/36 or 2/9. The proability of getting neither is equal to the probability of getting anything other than 7 or 8. We find this value by subtracting 2/9 from 1. So the probability of not getting 7 or 11 is 7/9.

5/12

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.

Find the probability of getting a sum of 7 and call that p(7). Now the complement rule says 1-p(7) is the probability of not getting 7.The ways we get 7 are 1 and 6, or 2 and 5, or 3 and 4.If you make a table with 1-6 on the top and 1-6 on the side, you can let each entry in your table be the sum of the top row and the side column. Since this is a 6x6 table, you can seethere are 36 outcomes.6 of those outcome result in a sum of 7.They are:1+66+12+55+23+44+3So p(7) =6/36 or 1/6. P of not getting 7 is 1-1/6=5/6

The probability is 1/6.

If the first roll is a three, it's impossible to get an eleven (the highest you can get is a 9), so the probability is 0. The way to get a sum of 7 is to add 3 and 4. The probability of getting a 4 on the second die is 1/6. Therefore, by adding the two probabilities of the outcomes (0 and 1/6), you get the answer: 1/6.

-- There are (6 x 6) = 36 possible rolls for a fair pair of 6-sided dice.-- There are 6 ways to roll a sum of 7 :1 ... 62 ... 53 ... 44 ... 35 ... 26 ... 1-- So the probability is 6/36 = 1/6 = 162/3 % .-- The probability is the probability, not the 'theoretical' probability.

Assuming they are standard and fair dice, the answer is 5/36+6/36 = 11/36.

The probability of getting a 7 on one roll of a die is zero.If you meant to ask about two dice, the probability is 6 in 36, or 1 in 6.

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