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)

_________________

Fabio Skilnik :: GMATH method creator (Math for the GMAT)

Our high-level "quant" preparation starts here: https://gmath.net