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The chances of occurrence of events are defined as a probability. The value of probability is always less than or equal to 1. Also, in the case of certain events, the probability is 1 and in the case of impossible events, the probability is 0.

Create the Sample Space for the experiment when two dice are rolled. When two dice are rolled, then there are 36 points in the sample space. So, the sample space for the given problem is as follows,

$\left(\begin{array}{cccccc}\left(1,1\right)& \left(1,2\right)& \left(1,3\right)& \left(1,4\right)& \left(1,5\right)& \left(1,6\right)\\ \left(2,1\right)& \left(2,2\right)& \left(2,3\right)& \left(2,4\right)& \left(2,5\right)& \left(2,6\right)\\ \left(3,1\right)& \left(3,2\right)& \left(3,3\right)& \left(3,4\right)& \left(3,5\right)& \left(3,6\right)\\ \left(4,1\right)& \left(4,2\right)& \left(4,3\right)& \left(4,4\right)& \left(4,5\right)& \left(4,6\right)\\ \left(5,1\right)& \left(5,2\right)& \left(5,3\right)& \left(5,4\right)& \left(5,4\right)& \left(5,6\right)\\ \left(6,1\right)& \left(6,2\right)& \left(6,3\right)& \left(6,4\right)& \left(6,5\right)& \left(6,6\right)\end{array}\right)$

Write the sample space with associated probabilities with each point for the sum obtained when two dice are rolled.

$\begin{array}{cccccccccccc}\begin{array}{l}Sample\\ Space\end{array}& 2& 3& 4& 5& 6& 7& 8& 9& 10& 11& 12\\ probability& \frac{1}{36}& \frac{2}{36}& \frac{3}{36}& \frac{4}{36}& \frac{5}{36}& \frac{6}{36}& \frac{5}{36}& \frac{4}{36}& \frac{3}{36}& \frac{2}{36}& \frac{1}{36}\end{array}$

It can be observed that there are 33 points in the sample space that gives the sum greater than or equal to 4, thus the probability is $\frac{33}{36}$or $\frac{11}{12}$.

Thus, the probability that the sum is greater than or equal to 4 is $\frac{11}{12}$.

Find the probability of getting an even sum by adding all the associated probabilities where the sum is even.

$\begin{array}{c}p=\frac{1}{36}+\frac{3}{36}+\frac{5}{36}+\frac{5}{36}+\frac{3}{36}+\frac{1}{36}\\ =\frac{18}{36}\\ =\frac{1}{2}\end{array}$

Thus, the probability that the sum is even is$\frac{1}{2}$ .

Find the probability of getting a sum which is divisible by 3 by adding all the associated probabilities where sum is divisible by 3.

$\begin{array}{c}p=\frac{2}{36}+\frac{5}{36}+\frac{4}{36}+\frac{1}{36}\\ =\frac{12}{36}\\ =\frac{1}{3}\end{array}$

Thus, the probability that the sum is divisible by 3 is$\frac{1}{3}$ .

Find the probability of getting a sum 7 from the obtained sample space.

$\begin{array}{c}p=\frac{6}{36}\\ =\frac{1}{6}\end{array}$

Thus, the probability of getting a sum 7 is $\frac{1}{6}$.

The product of two number is 12 when the pairs of numbers are $\left(2,6\right)$and $\left(3,\text{鈥�}4\right)$.Thus the points in the sample space which gives 12 as the product are $\left(2,6\right)\text{鈥�},\left(6,2\right)\text{鈥�},\left(3,4\right)\text{鈥�},\left(4,3\right)$and each has a probability of $\frac{1}{36}$.

Find the probability of getting a product 12 by adding the probability of each point.

$\begin{array}{c}p=\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}\\ =\frac{4}{36}\\ =\frac{1}{9}\end{array}$

Thus, the required probability is $\frac{1}{9}$.