Bunuel wrote:
The probability of winning game A is X and the probability of winning game B is Y. When playing a single round of each of the two games, what is the probability of winning exactly once?
A. X-YX
B. XY
C. X+Y
D. X+Y-2XY
E. X^2Y^2-XY^2-X^2Y
\(? = P\left( {{\text{win}}\,\,{\text{just}}\,\,{\text{A}}\,\,{\text{or}}\,\,{\text{win}}\,\,{\text{just}}\,\,{\text{B}}} \right)\,\,\,\, = \,\,\,\,P\left( {{\text{win}}\,\,{\text{A}}\,{\text{or}}\,\,{\text{B}}} \right) - P\left( {{\text{win}}\,\,{\text{both}}} \right)\)
\(P\left( {{\rm{win}}\,\,{\rm{both}}} \right) = xy\)
\(\left[ {\,x,y\,\,{\rm{constants}}\,\,\,\, \Rightarrow \,\,\,\,{\rm{win}}\,A\,\,{\rm{and}}\,\,{\rm{win}}\,B\,\,{\rm{are}}\,\,{\rm{independent}}\,\,{\rm{events}}\,} \right]\)
\(P\left( {{\text{win}}\,\,{\text{A}}\,{\text{or}}\,\,{\text{B}}} \right) = x + y - xy\,\,\,\,\,\left[ {\,{\text{simplifier}}\,} \right]\)
\(? = x + y - 2xy\)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
"Kolmogorov - Poincaré - Gauss - Euler - Newton, are only five lives separating us from the source of our science."
(Vladimir Arnold)
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