Conditional Probability Problem?

You draw two cards from a well-shuffled deck of 52 cards. After you draw the first card, you do not replace it in the deck. What is the probability of P(get ace of diamonds on first draw and get king of clubs on second draw)?

A) 2/51
B) 1/26
C) 1/2652
D) 1/2704

Question

Conditional Probability Problem?

You draw two cards from a well-shuffled deck of 52 cards. After you draw the first card, you do not replace it in the deck. What is the probability of P(get ace of diamonds on first draw and get king of clubs on second draw)? 

A) 2/51 
B) 1/26 
C) 1/2652 
D) 1/2704

anonymous · Answer

Let's call A the event that you get an ace of diamonds, and B the event that you get a king of clubs.

You want to find P(A and B).

Use the formula for conditional probability:
P(B|A) = P(A and B) / P(A)

P(A) is fairly obvious, it must be 1/52.

P(B|A) is pretty easy to find, too. There's only one king of clubs, and since the ace has already been drawn, there is a total of only 51 cards left. This means the probability of drawing a king of clubs (after having drawn an ace of diamonds) is 1/51.

1/51 = P(A and B) / (1/52), or
1/(51*52) = P(A and B) = ∙∙∙