Question 401494
Number of Favourable/Favorable Choices 	= The number of ways in which one {{{ace}}} can be drawn from the total {{{4}}}

 {{{mA}}} 	= 	{{{4C1}}}

	= 	{{{4/1}}}

	= 	{{{4}}}


Probability of the card drawn being an {{{ace}}}

Probability of occurrence of Event "A" = 	
{{{(Number_ of_ Favourable_ Choices_ for_ the_ Event)}}}/
{{{Total_ Number_ of_ Possible_ Choices _for_ the_ Experiment}}}


 {{{P(A)}}} 	= 	{{{mA/n}}}

	= 	{{{4/52}}}

	= 	{{{1/13}}}


so, your answer is: {{{1/13}}}