Sum Rule:
P(A) = \sum_{B} P(A,B)
Product Rule:
P(A , B) = P(A) P(B|A) or P(A, B)=P(B) P(A|B)
[P(A|B) means probability of A given that B has occurred]
P(A, B) = P(A) P(B) , if A and B are independent events.
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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.
The probability is 1.
The sum should equal to 1.
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.
If you rolled 2 fair dice, the probability of having a sum of 6 is 5 over 36