When a single card is drawn from an ordinary 52 card deck, find 
the probability it will be the following; 

A red Card:  There are 26 red cards out of 52 cards, so the 
probability is P(R) = 26/52 = 1/2

An Ace:  There are 4 aces out of 52 cards, so the probability is
P(A) = 4/52 = 1/13

A heart or a king:  P(H or K) = P(H)+P(K)-P(H and K)

There are 13 hearts out of 52 cards, so P(H) = 13/52 = 1/4
There are 4 kings out of 52 cards, so P(K) = 4/52 = 1/13
There is 1 card which is a heart and a king (the king of 
hearts), so P(H and K) = 1/52, so

P(H or K) = P(H)+P(K)-P(H and K) = 1/4+1/13-1/52 = 
13/52+4/52-1/52 = 16/52 = 4/13

A club, given it is black: P(C|B) = P(C and B)/P(B)

There are 13 cards out of 52 which Clubs and Blacks so 
P(C and B) = 13/52 = 1/4
There are 26 cards out of 52 which are black, so
P(B) = 26/52 = 1/2
P(C|B) = P(C and B)/P(B) = (1/4)/(1/2) = (1/4)(2/1) = 1/2

An ace or a card above a jack

Hmmm!  This will depend on whether and ace is considered 
higher or lower than a jack.  It is usually higher, but 
sometimes it's lower. I'll do both cases:

Case 1.  An ace is considered higher than a jack:

P(A or Above Jack) = P(A) + P(Above Jack) - P(A and Above Jack)
There are 4 aces, so P(A) = 4/52 = 1/13 
There are 4 queens, 4 kings and 4 aces which are above jack, so
P(Above jack) = 12/52 = 3/13
All 4 aces are above Jack so P(A and Above Jack) = 4/52 = 1/13, so
P(A or Above Jack) = P(A) + P(Above Jack) - P(A and Above Jack) =
1/13 + 3/13 - 1/13 = 3/13

Case 2.  An ace is considered lower than a jack:

P(A or Above Jack) = P(A) + P(Above Jack) - P(A and Above Jack)
There are 4 aces, so P(A) = 4/52 = 1/13 
There are 4 queens and 4 kings which are above jack, so
P(Above jack) = 8/52 = 2/13
None of the 4 aces are above Jack so P(A and Above Jack) = 0/52 =
0, so 
P(A or Above Jack) = P(A) + P(Above Jack) - P(A and Above Jack) =
1/13 + 3/13 - 0 = 4/13

A queen, given it is a spade: P(Q|S) = P(Q and S)/P(S)
There is only one card that is a queen and a spade, namely the 
queen of spades, so P(Q and S) = 1/52. There are 13 spades, 
so P(S) = 13/52 = 1/4, so
P(Q|S) = P(Q and S)/P(S) = (1/52)/(1/4) = (1/52)(4/1) = 1/13

A 5, given it is a face card

P(5|F) = P(5 and F)/P(F)
Since a 5 is not a face card, P(5 and F) = 0
Since there are 4 Jacks, 4 queens, and 4 kings which are face 
cards, P(F) = 12/52 = 3/13, so
P(5|F) = P(5 and F)/P(F) = 0/(3/13) = 0(13/3) = 0
---------------
Now assume that the two cards are drawn without replacement. 
Find the probability of each of the following events. 

Both are aces:
P(A 1st and A 2nd) = P(A 1st)·P(A 2nd) = (4/52)(3/51) = 1/221 

One is a king and one is an ace:
P[(A 1st and K 2nd) or (K 1st and A 2nd)] = 
P(A 1st and K 2nd) + P(K 1st and A 2nd) =
P(A 1st)·P(K 2nd) + P(K 1st)·P(A 2nd) =
(4/52)(4/51) + (4/52)(4/51) = 8/663

Neither one is an ace:
There are 48 non-aces, so P(non-A 1st and non-A 2nd) =
P(non-A 1st)·P(non-A 2nd) = (48/52)(47/51) = 188/221 

Both are red:
P(R 1st and R 2nd) = P(R 1st)·P(R 2nd) = (26/52)(25/51) = 
25/102

Both are face cards:
P(F 1st and F 2nd) = P(F 1st)·P(F 2nd) = (12/52)(11/51) = 
11/221

A face card
Here it is easier to use the complement event, 
P(one is a face card) = 1 - P(both are non-face card). 
(There are 12 face cards and 40 non-face cards)

1 - P(non-F 1st and non-F 2nd) = 1 - P(non-F 1st)·P(non-F 2nd) =
1 - (40/52)(39/51) = 7/17 

This could have also been worked a longer way without using the 
complement event:

P[(F 1st and F 2nd) or (F 1st and non-F 2nd) or (non-F 1st and F 2nd)] =
P(F 1st and F 2nd) + P(F 1st and non-F 2nd) + P(non-F 1st and F 2nd) =
P(F 1st)·P(F 2nd) + P(F 1st)·P(non-F 2nd) + P(non-F 1st)·P(F 2nd) =
(12/52)(11/51) + (12/52)(40/51) + (40/52)(12/52) =
11/221 + 40/221 + 40/221 = 7/17

A black 2 or a face card
This one is also much easier to work using the complement event:
There are 2 black 2's and 12 face cards or 14 which are one or the other.
That leaves 38 cards which are neither.

P(B2 or F) = 1 - P(neither 1st and neither 2nd) =
P(neither 1st)·P(neither 2nd) = (38/52)·(37/51) = 703/1326 

This could have also been done a much longer way without using the
complement event, but I will not bother with that.

Edwin