Are the following correct non-uniform sample spaces for a throw of two dice? If

so, find the probabilities of the given sample points. If not show what is wrong.

Suggestion: Copy sample space (2.4) and circle on it the regions corresponding to the points of the proposed non-uniform spaces.

(a) First die shows an even number.

First die shows an odd number.

(b) Sum of two numbers on dice is even.

First die is even and second odd.

First die is odd and second even.

(c) First die shows a number≤3.

At least one die shows a number > 3.

Non-Uniform Sample spaceof any experiment is the set of all possible mutually exclusive events such that each point has different probability.

When two die are rolled, there are 36 points in the sample space.

Find the sample space for the given problem.

$$\begin{gathered} 1,11,21,31,41,51,6 \\ 2,12,22,32,42,52,6 \\ 3,13,23,33,43,53,6 \\ 4,14,24,34,44,54,6 \\ 5,15,25,35,45,55,6 \\ 6,16,26,36,46,56,6 \end{gathered}$$

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

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

$$\begin{gathered} \frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{3}{9} \\ =\frac{1}{3} \end{gathered}$$

Find the probability that first die is even and second odd.

$\frac{9}{36}=\frac{1}{4}$

Find the probability that first die is odd and second even.

$\frac{9}{36}=\frac{1}{4}$

It can be observed that the probabilities of each sample point are not equal, thus the sample space is non-uniform.

Find the probability that First die shows a number is less than or equal to 3.

$\frac{18}{36}=\frac{1}{2}$

Find the probability that at least one die shows a number is greater than 3.

$\frac{27}{36}=\frac{3}{4}$

It can be observed that the probabilities of each sample point are not equal, thus the sample space is non-uniform.