91Ó°ÊÓ

The uniform sample space of an experiment is the set of outcomes having the similar probability of occurring.

The non-uniform sample space of an experiment is the set of all possible mutually exclusive events that is each point has a different probability.

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,

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

It can be observed from the sample space that there are 21 points in the sample space that gives a number greater than 33, thus the probability that a two-digit number is greater than 33 with the two numbers on the dice is formed asor $\frac{21}{36}$or $\frac{7}{12}$.

Thus, the required probability is $\frac{7}{12}$.

It can be observed that from the sample space there are 17 points in the sample space that gives a number greater than or equal to 42, thus the probability that a two-digit number is greater than or equal to 42 with the two numbers on the dice is formed is $\frac{17}{36}$.

Thus, the required probability is$\frac{17}{36}$ .

It can be observed from the sample space that all numbers except $\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(4,4\right),\left(5,5\right),\left(6,6\right)$can form a larger number as well as a smaller number with the same probability.

Thus, the required numbers are other than $\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(4,4\right),\left(5,5\right),\left(6,6\right)$.