Question 171510
First of all, the statement describing the experiment is ambiguous.  Are you drawing one card and it must be exactly the 5 of spades, or are you drawing two cards, one of which must be a 5 and the other must be a spade?  Furthermore,  you don't specify if, on a two card draw, whether you are replacing the first card before making the second draw.


Case 1:  One card draw and success = 5 of spades. Trivial.  There is only one 5 of spades in the deck, so your probability is {{{1/52}}}


Case 2:  Two card draw with replacement and success = one card will be a spade of any rank and the other will be a 5 of any suit.

First of all, there are 13 spades out of 52 cards, so the probability of drawing any spade is {{{13/52}}}.


Second, there are four 5s in the deck, so the probability of drawing a 5 is {{{4/52}}}


And the total probability is the product {{{13/52*4/52=52/2704}}}, approximately 0.019 or 1.9%


Case 3: Two card draw WITHOUT replacement and success is the same as Case 2.


You have to consider the possibility that the first card drawn is the 5 of spades. So:


Case 3a: The probability of drawing a spade other than the 5 times the probability of drawing a 5 when the deck is one card smaller {{{12/52*4/51}}}


Plus


Case 3b: The probability of drawing the 5 of spades times the probability of drawing a 5 with a deck that is one card smaller AND has one less 5 {{{1/52*3/51}}}


Total probability {{{(12/52*4/51)+(1/52*3/51)}}} roughly the same value as Case 2.


Case 4:  You really meant to ask "What is the probability, on a one card draw, that the card will be a spade OR a 5"


Again 13 spades and 4 fives one of which is a spade and already counted, so there are 3 other suited fives, making a total of 16 cards that represent a successful experiment, and your probability is {{{16/52}}}  (You could also look at it as 4 fives and 12 spades that aren't a five adding up to the same 16 successes out of 52 possibilities.)