Probability question: Cards

Question

Suppose that you select 2 cards without replacement from a standard deck of 52 playing cards.  If the first card that you select is a club, what is the probability that the second card that you select is a five?

Answer 1

Two events must be considered.

 

Event 1:The first card is a club but is not the five of clubs.  The second card is a five.  Denote its probability by P(CN5,5).

 

Event 2: The first card is the five of clubs.  The second card is a five.  Denote its probability by P(C5,5).

 

These two events are mutually exclusive, hence  The probability that the second card is a five is P(CN5,5)+P(C5,5).  Denote this probability by P(C,5)

 

There are 12 clubs that are not a five.  There is one club that is a five.

 

Probability that the first card is a club but not a five = 12/52.  Denote this probability by P(CN5).

 

Probability that the first card is the five of clubs = 1/52.  Denote this probability by P(C5).

 

The probability that the second card is a five if the first card is not = 4/51.  Denote this probability by P(5|CN5).

 

The probability that the second card is a five if the first card is the five of clubs = 3/51.  Denote this probability by P(5|C5).

 

If an event A can come about through two mutually exclusive ways, B and C, then P(A)=P(B)P(A|B)+P(C)P(A|C).  Applying this rule:

 

Denote the probability that the second card is a five by p(5).

 

P(C,5)=P(CN5)P(5|CN5)+P(C5)P(5|C5)

 

P(C,5)=(12/52)(4/51)+(1/52)(3/51)=(12*4+1*3)/(52*51)=51/(52*51)=1/52

 

Check the work to make sure I didn't make any errors.