The probability of winning game A is X and the probability

BTGmoderatorDC 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

Source: Economist GMAT

$$? = P\left( {{\text{win}}\,\,{\text{just}}\,\,{\text{A}}\,\,{\text{or}}\,\,{\text{win}}\,\,{\text{just}}\,\,{\text{B}}} \right)\,\,\,\, = \,\,\,\,P\left( {{\text{win}}\,\,{\text{A}}\,{\text{or}}\,\,{\text{B}}} \right) - P\left( {{\text{win}}\,\,{\text{both}}} \right)$$
$$P\left( {{\rm{win}}\,\,{\rm{both}}} \right) = xy\,\,\,\,\,\,\,\,\left[ {\,x,y\,\,\,{\rm{constants}}\,\,\,\, \Rightarrow \,\,\,\,{\rm{independency}}\,} \right]$$
$$P\left( {{\text{win}}\,\,{\text{A}}\,{\text{or}}\,\,{\text{B}}} \right) = x + y - xy\,\,\,\,\,\left[ {\,{\text{simplifier}}\,} \right]$$
$$? = x + y - 2xy$$



This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.


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