BTGmoderatorDC wrote:In a certain game, you perform three tasks. You flip a quarter, and success would be heads. You roll a single die, and success would be a six. You pick a card from a full playing-card deck, and success would be picking a spades card. If any of these task are successful, then you win the game. What is the probability of winning?
A. 1/48
B. 5/16
C. 11/12
D. 11/16
E. 23/48
Source: Magoosh
$$\left. \matrix{
{\rm{i}}\,\,{\rm{:}}\,\,\,{\rm{get}}\,\,{\rm{H}}\,\,{\rm{in}}\,\,{\rm{quarter}} \hfill \cr
{\rm{ii}}:\,\,{\rm{get}}\,\,6\,\,{\rm{in}}\,{\rm{die}} \hfill \cr
{\rm{iii}}\,\,{\rm{:}}\,\,{\rm{get}}\,\,{\rm{spades}}\,{\rm{in}}\,{\rm{52}}\,\,{\rm{cards}}\,{\rm{regular}}\,\,{\rm{deck}} \hfill \cr} \right\}\,\,\,{\rm{independent}}\,\,{\rm{events}}\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{their}}\,\,{\rm{negations}}\,\,{\rm{are}}\,\,{\rm{independent}}\,\,{\rm{events}}\,\,\,\,\,\left( * \right)$$
$$? = P\left( {{\rm{i}}\,\,{\rm{OR}}\,\,{\rm{ii}}\,\,{\rm{OR}}\,\,{\rm{iii}}} \right) = 1 - P\left( {\left( {{\rm{not}}\,\,{\rm{i}}} \right)\,\,AND\,\,\left( {{\rm{not}}\,\,{\rm{ii}}} \right)\,\,AND\,\,\left( {{\rm{not}}\,\,{\rm{iii}}} \right)} \right)$$
$$P\left( {\left( {{\rm{not}}\,\,{\rm{i}}} \right)\,\,AND\,\,\left( {{\rm{not}}\,\,{\rm{ii}}} \right)\,\,AND\,\,\left( {{\rm{not}}\,\,{\rm{iii}}} \right)} \right)\,\,\mathop = \limits^{\left( * \right)} \,\,\,{1 \over 2} \cdot {5 \over 6} \cdot {{52 - 13} \over {52}} = {5 \over {16}}$$
$$? = 1 - {5 \over {16}} = {{11} \over {16}}$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
