Listing a sample space. The sample space \(\displaystyle S\) of an experiment is the set of
all the outcomes it can produce, with nothing left out and nothing counted twice. When an experiment has two parts done together, each outcome has to record both parts.
(i) Rolling a die and tossing a coin together. One outcome must say what the die showed
and what the coin showed, so write it as an ordered pair (die, coin). The die contributes $\displaystyle 6$ possibilities and the coin $\displaystyle 2$, and every die result can pair with either coin result:
\[S=\{(1,H),(2,H),(3,H),(4,H),(5,H),(6,H),(1,T),(2,T),(3,T),(4,T),(5,T),(6,T)\} \]
\[n(S)=6\times 2=12 \]
The multiplication \(\displaystyle 6\times 2\) is just a shortcut for the counting you can see in the list: $\displaystyle 6$ outcomes with \(\displaystyle H\), then the same $\displaystyle 6$ again with \(\displaystyle T\).
(ii) Choosing a random integer between \(\displaystyle -5\) and \(\displaystyle +5\). Read strictly, "between \(\displaystyle -5\) and \(\displaystyle +5\)" means the two end numbers are not themselves included, so the integers available are
\[S=\{-4,-3,-2,-1,0,1,2,3,4\},\qquad n(S)=9 \]
Do not forget \(\displaystyle 0\) - it is an integer, and it lies between \(\displaystyle -5\) and \(\displaystyle 5\). (A word of caution: the word "between" is used loosely in everyday speech. If your teacher intends the endpoints to be included, the sample space is \(\displaystyle S=\{-5,-4,-3,-2,-1,0,1,2,3,4,5\}\) with \(\displaystyle n(S)=11\). Say which reading you are using, and either is defensible; the strict reading, giving $\displaystyle 9$, is the one taken here.)
(iii) Drawing one ball from a box of $\displaystyle 5$ green and $\displaystyle 7$ red balls. This depends on what you decide to record.
If you only record the colour*, then \(\displaystyle S=\{\text{green},\ \text{red}\}\) and \(\displaystyle n(S)=2\). But be careful - these two outcomes are
not equally likely, so you must not compute probabilities by dividing by 2.
To get a sample space of equally likely* outcomes, treat the balls as distinguishable - imagine numbering them:
\[S=\{G_1,G_2,G_3,G_4,G_5,\ R_1,R_2,R_3,R_4,R_5,R_6,R_7\},\qquad n(S)=5+7=12 \]
Each of these $\displaystyle 12$ balls is equally likely to be the one drawn, and now counting works:
\[P(\text{green})=\frac{5}{12},\qquad P(\text{red})=\frac{7}{12},\qquad \frac{5}{12}+\frac{7}{12}=1\ \checkmark \]
Answers: (i) \(\displaystyle S=\{(1,H),(2,H),(3,H),(4,H),(5,H),(6,H),(1,T),(2,T),(3,T),(4,T),(5,T),(6,T)\}\), \(\displaystyle n(S)=12\); (ii) \(\displaystyle S=\{-4,-3,-2,-1,0,1,2,3,4\}\), \(\displaystyle n(S)=9\) (or \(\displaystyle n(S)=11\) if \(\displaystyle -5\) and \(\displaystyle +5\) are meant to be included); (iii) by colour \(\displaystyle S=\{\text{green},\text{red}\}\) with $\displaystyle 2$ unequally likely outcomes, or, taking the balls as distinct, \(\displaystyle S=\{G_1,\dots,G_5,R_1,\dots,R_7\}\) with \(\displaystyle n(S)=12\) equally likely outcomes.