Bunuel wrote:
Pam and Robin each roll a pair of fair, six-sided dice. What is the probability that Pam and Robin will both roll the same set of two numbers?
A. 1/216
B. 1/36
C. 5/108
D. 11/216
E. 1/18
\(? = {{\# \,\,{\rm{favorables}}} \over {\# \,\,{\rm{total}}\,\,\left( {{\rm{equiprobables}}} \right)}}\)
\(\# \,\,{\rm{total}}\,\,\left( {{\rm{equiprobables}}} \right) = {6^2} \cdot {6^2}\,\,\,\,\left( {{\rm{taking}}\,\,{\rm{results}}\,\,{\rm{in}}\,\,{\rm{order,}}\,\,{\rm{for}}\,\,{\rm{both}}\,\,{\rm{players}}} \right)\)
\(\# \,\,{\rm{favorables}}\,\,\, = \,\,\,\left\{ \matrix{\\
\,6\,\,::\,\,{\rm{first}}\,\,{\rm{player}}\,\,{\rm{gets}}\,\,{\rm{same}}\,\,{\rm{number}}\,\,{\rm{twice,}}\,\,{\rm{second}}\,\,{\rm{player}}\,\,{\rm{gets}}\,\,{\rm{same}}\,\,{\rm{ones}} \hfill \cr \\
\,\,\, + \,\,\,\,\,\left( {{\rm{mutually}}\,\,{\rm{exclusive}}\,{\rm{!}}} \right) \hfill \cr \\
\,\left( {36 - 6} \right) \cdot 2!\,\,\,::\,\,\,{\rm{first}}\,\,{\rm{gets}}\,\,\left( {x,y} \right)\,\,{\rm{with}}\,\,x \ne y\,\,,\,\,{\rm{second}}\,\,{\rm{gets}}\,\,\left( {x,y} \right)\,\,{\rm{or}}\,\,\left( {y,x} \right) \hfill \cr} \right.\)
\(? = {{66} \over {{6^4}}} = {{11} \over {{6^3}}} = {{11} \over {216}}\)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.