A die is tossed twice. What is the probability of obtaining a total of 6 if both numbers are odd?
Explanation
For the 6 faces die tossed twice, we construct the probability table as shown
\begin{array}{|c|c|c|c|c|c|c|}
\hline\(+\) & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline 1 & 1,1 & 1,2 & 1,3 & 1,4 & 1,5 & 1,6 \\
\hline 2 & 2,1 & 2,2 & 2,3 & 2,4 & 2,5 & 2,6 \\
\hline 3 & 3,1 & 3,2 & 3,3 & 3,4 & 3,5 & 3,6 \\
\hline 4 & 4,1 & 4,2 & 4,3 & 4,4 & 4,5 & 4,6 \\
\hline 5 & 5,1 & 5,2 & 5,3 & 5,4 & 5,5 & 5,6 \\
\hline 6 & 6,1 & 6,2 & 6,3 & 6,4 & 6,5 & 6,6 \\
\hline
\end{array}
Pr (a total ot 6 and both odd) \(=\frac{\text { event space }}{\text { sample space }}\)vent space \(=e\) (5.1). (3.3).
\(\operatorname{Pr}(\) a total of 6 and both odd)
$$
=\frac{\text { event space }}{\text { sample space }}
$$
\(n(\) event space \()=3, n(\) sample space \()=36\)
\(\operatorname{Pr}\left(\right.\) a total of 6 and both odd) \(=\frac{3}{36}=\frac{1}{12}\)