Two fair dice are tossed, and the following events are defined:

A: {Sum of the numbers showing is odd.}

B: {Sum of the numbers showing is 9, 11, or 12.}

Are events A and B independent? Why?

The primary outcomes of an experiment are the sample points in the sample space. The experiment is represented in terms of the sample space by sample points, which are sample space elements. Sampling units or observations are other terms for sample points.

Rolling a two fair dice gives the following sample points:

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Therefore, the sample points are:

A: {(1, 2), (1, 5), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5)}

B: {(3, 6), (4, 5), (5, 4), (5, 6), (6, 3), (6, 5), (6, 6)}

$\mathbf{\text{A}}\mathbf{\cap }\mathbf{\text{B:}}\mathbf{\left\{}\left(\text{3, 6}\right)\mathbf{\text{,}}\left(\text{4, 5}\right)\mathbf{\text{,}}\left(\text{5, 4}\right)\mathbf{\text{,}}\left(\text{5, 6}\right)\mathbf{\text{,}}\left(\text{6, 3}\right)\mathbf{\text{,}}\left(\text{6, 5}\right)\mathbf{\right\}}$

The probabilities are:

$\begin{array}{l}\mathbf{\text{P (A) =}}\frac{\mathbf{\text{18}}}{\mathbf{\text{36}}}\\ \mathbf{\text{P (B) =}}\frac{\mathbf{\text{7}}}{\mathbf{\text{36}}}\\ \mathbf{\text{P (C) =}}\frac{\mathbf{\text{6}}}{\mathbf{\text{36}}}\end{array}$

For independent

$\begin{array}{c}\mathbf{\text{P (A}}\mathbf{\cap }\mathbf{\text{B) = P (A)}}\mathbf{\times }\mathbf{\text{P(B)}}\\ \frac{\mathbf{\text{6}}}{\mathbf{\text{36}}}\mathbf{\text{=}}\frac{\mathbf{\text{18}}}{\mathbf{\text{36}}}\mathbf{\times }\frac{\mathbf{\text{7}}}{\mathbf{\text{36}}}\\ \frac{\mathbf{\text{1}}}{\mathbf{\text{6}}}\mathbf{\text{=}}\frac{\mathbf{\text{126}}}{\mathbf{\text{1296}}}\\ \frac{\mathbf{\text{1}}}{\mathbf{\text{6}}}\mathbf{\ne }\frac{\mathbf{\text{7}}}{\mathbf{\text{72}}}\end{array}$

Hence,$\mathbf{\text{P (A}}\mathbf{\cap }\mathbf{\text{B)}}\mathbf{\ne }\mathbf{\text{P (A)}}\mathbf{\times }\mathbf{\text{P(B)}}$ it means A and B are not independent.