Refer to Example 13. (i) Complete the following table (ii) 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 $\frac{1}{11}$. Do you agree with this argument? Justify

Solution: When two dice are rolled, then their sum can be between $2$ to $12$.
Here, total number of possible outcomes,$n\left(S\right)=6\ast 6=36$

When sum is 2, possible outcomes are,
$S_2=\left(1,1\right)$
So,$P\left(S_2\right)=\frac{1}{36}$

When sum is 3, possible outcomes are,
$S_3=\left\{\left(1,2\right),\left(2,1\right)\right\}$
So,$P\left(S_3\right)=\frac{2}{36}=\frac{1}{18}$

When sum is 4, possible outcomes are,
$S_4=\left\{\left(1,3\right),\left(3,1\right),\left(2,2\right)\right\}$
So,$P\left(S_4\right)=\frac{3}{36}=\frac{1}{12}$

When sum is 5, possible outcomes are,
$S_5=\left\{\left(1,4\right),\left(4,1\right),\left(2,3\right),\left(3,2\right)\right\}$
So,$P\left(S_5\right)=\frac{4}{36}=\frac{1}{9}$

When sum is 6, possible outcomes are,
$S_6=\left\{\left(1,5\right),\left(5,1\right),\left(3,3\right),\left(4,2\right),\left(2,4\right)\right\}$
So,$P\left(S_6\right)=\frac{5}{36}$

When sum is 7, possible outcomes are,
$S_7=\left\{\left(1,6\right),\left(6,1\right),\left(4,3\right),\left(5,2\right),\left(2,5\right),\left(3,4\right)\right\}$
So,$P\left(S_7\right)=\frac{6}{36}=\frac{1}{6}$

When sum is 8, possible outcomes are,
$S_8=\left\{\left(2,6\right),\left(6,2\right),\left(5,3\right),\left(3,5\right),\left(4,4\right)\right\}$
So,$P\left(S_8\right)=\frac{5}{36}$

When sum is 9, possible outcomes are,
$S_9=\left\{\left(3,6\right),\left(6,3\right),\left(5,4\right),\left(4,5\right)\right\}$
So,$P\left(S_9\right)=\frac{4}{36}=\frac{1}{9}$

When sum is 10, possible outcomes are,
$S_{10}=\left\{\left(4,6\right),\left(6,4\right),\left(5,5\right)\right\}$
So,$P\left(S_{10}\right)=\frac{3}{36}=\frac{1}{12}$

When sum is 11, possible outcomes are,
$S_{11}=\left\{\left(5,6\right),\left(6,5\right)\right\}$
So,$P\left(S_{11}\right)=\frac{2}{36}=\frac{1}{18}$

When sum is 12, possible outcomes are,
$S_{12}=\left\{\left(6,6\right)\right\}$
So,$P\left(S_{12}\right)=\frac{1}{36}$

So, with the above values we can complete the given table.

(ii) As the student argues that probability of each sum is same, he is incorrect as we have shown above each of these values are not identical and not equal to $\frac{1}{11}$.