Poker card problem

[Elix]

This is the question I'm struggling with: You draw two cards, without replacement, from a well-shuffled, standard poker deck. What is the chance that one card will be a spade? (exactly one). This is what I've tried so far:

1st combination:
(Probability of 1st card being a spade) = 13/52
(Prob. of the 2nd card not being a spade) = 39/51

2nd combination:
(Prob. of the 2nd card not being a spade) = 39/51
(Probability of 1st card being a spade) = 13/52

Work:
[MATH]13/52 * 39/51 = 13/68 [/MATH][MATH] 13/68 + 13/68 = 26/136 [/MATH]

 

[Romsek]

this is correct except for a silly addition error at the end.

An easier and more fundamental way to view this is

\(\displaystyle p[\text{exactly 1 spade in two draws}] = \dfrac{\dbinom{13}{1}\dbinom{39}{1}}{\dbinom{52}{2}} = \dfrac{13}{34}\)

 

[Elix]

this is correct except for a silly addition error at the end.

An easier and more fundamental way to view this is

\(\displaystyle p[\text{exactly 1 spade in two draws}] = \dfrac{\dbinom{13}{1}\dbinom{39}{1}}{\dbinom{52}{2}} = \dfrac{13}{34}\)

Ohh I see now.. thanks a bunch for pointing that out!

 

[Steven G]

This is the question I'm struggling with: You draw two cards, without replacement, from a well-shuffled, standard poker deck. What is the chance that one card will be a spade? (exactly one). This is what I've tried so far:

1st combination:
(Probability of 1st card being a spade) = 13/52
(Prob. of the 2nd card not being a spade) = 39/51

2nd combination:
(Prob. of the 2nd card not being a spade) = 39/51
(Probability of 1st card being a spade) = 13/52

Work:
[MATH]13/52 * 39/51 = 13/68 [/MATH][MATH] 13/68 + 13/68 = 26/136 [/MATH]

Your 1st and 2nd combinations are the same!

You are really going to have a hard time in probability if you do not know how to add fractions