cnon wrote:

Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 competitors in the triathlon, and medals are awarded for first, second, and third place, what is the probability that at least two of the triplets will win a medal?

A. 3/14

B. 19/84

C. 11/42

D. 15/28

E. 3/4

\(\left\{ \matrix{

\,A\,,B\,,\,{\kern 1pt} C\,\,\,\,\, \to \,\,\,3\,\,{\rm{competitors}} \hfill \cr

\,{\rm{D,}}\,\, \ldots \,\,{\rm{,}}\,I\,\,\, \to \,\,\,{\rm{other}}\,\,6\,\,{\rm{competitors}}\, \hfill \cr} \right.\)

\(\left( {\rm{i}} \right)\,\,A,B,C\,\,{\rm{win}}\,\,{\rm{the}}\,\,3\,\,{\rm{medals}}\)

\(\left( {{\rm{ii}}} \right)\,\,A,B,C\,\,{\rm{win}}\,\,{\rm{exactly}}\,\,2\,\,{\rm{of}}\,\,{\rm{the}}\,\,3\,\,{\rm{medals}}\)

\(? = P\left( {{\rm{i}}\,\,{\rm{or}}\,\,{\rm{ii}}} \right) = P\left( {\rm{i}} \right) + P\left( {{\rm{ii}}} \right)\,\,\,\,\,\,\,\,\left[ {\,{\rm{i}}\,\,{\rm{and}}\,\,{\rm{ii}}\,\,{\rm{are}}\,\,{\rm{mutually}}\,\,{\rm{exclusive}}\,} \right]\)

\(\left( {\rm{i}} \right)\,\,\,\left\{ \matrix{

\,{\rm{favorable}} = \,\,3! \hfill \cr

\,{\rm{total}}\,\, = \,\,9 \cdot 8 \cdot 7\,\,{\rm{equiprobables}}\,\,\, \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,P\left( {\rm{i}} \right) = {1 \over {84}}\)

\(\left( {{\rm{ii}}} \right)\,\,\,\left\{ \matrix{

\,{\rm{favorable}} = \,\,\underbrace {C\left( {3,2} \right)}_{{\rm{say}}\,\,A,B}\,\,\, \cdot \,\,\underbrace {C\left( {3,2} \right)}_{{\rm{say}}\,\,\,1{\rm{st}}\,\,,\,\,2{\rm{nd}}}\,\,\, \cdot \,\,\underbrace {2!}_{{\rm{say}}\,\,\,A\,\,1{\rm{st}}\,\,,\,B\,\,2{\rm{nd}}}\,\,\, \cdot \,\,\,\underbrace {C\left( {6,1} \right)}_{{\rm{say}}\,\,D\,\,\left( {3{\rm{rd}}} \right)}\, \hfill \cr

\,{\rm{total}}\,\, = \,\,9 \cdot 8 \cdot 7\,\,{\rm{equiprobables}}\,\,\, \hfill \cr} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,P\left( {{\rm{ii}}} \right) = {3 \over {14}}\)

\(? = {1 \over {84}} + {{3 \cdot 6} \over {14 \cdot 6}} = {{19} \over {84}}\)

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

Regards,

Fabio.

_________________

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

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