Two dice,one blue and one grey, are thrown at the same time. Complete the following table: Event: Sum on two dice, 2, 3, 4, 5, 6, 7, 8, 9,10, 11, 12 Probability_,_ ,_ ,_ ,_ ,_ ,_ ,_ , ,_ ,_ ,_ From the above table a student argues that there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability 1/11. Do you agree with this argument?

Explanation

When two dice are rolled, then their sum can be between `2` to `12`.

Here, total number of possible outcomes,`n(S) = 6**6 =36`


When sum is 2, possible outcomes are,

`S_2 = (1,1)`

So,`P(S_2) = 1/36`


When sum is 3, possible outcomes are,

`S_3 = {(1,2),(2,1)}`

So,`P(S_3) = 2/36 = 1/18`


...