# 31.14 - Playing cards

• May 8th 2008, 08:27 AM
ccdelia7
31.14 - Playing cards
Two cards are sequentially drawn (without replacement) from a standard deck of 52 cards (well shuffled). Let A be the event that the two cards have the same value. Let B be the event that the first card drawn is an Ace. Are these events independent?

Definition: Two events are said to be "independent" provided that P(A intersection B) = P(A)P(B).
• May 8th 2008, 08:38 AM
janvdl
Quote:

Originally Posted by ccdelia7
Two cards are sequentially drawn (without replacement) from a standard deck of 52 cards (well shuffled). Let A be the event that the two cards have the same value. Let B be the event that the first card drawn is an Ace. Are these events independent?

Definition: Two events are said to be "independent" provided that P(A intersection B) = P(A)P(B).

For number 1:

Well the first card can be any card... The second card just needs to be the same value as the first one.

So for example, if you drew one of the 4 Aces, then there are 3 Aces and 51 cards in total remaining. You thus have a $\frac{3}{51}$ chance of drawing the same value card again.

I'll think a bit more about number 2.
• May 8th 2008, 08:41 AM
Moo
Number two..

P(A intersection B)=probability that the first card is an ace AND the first two cards have the same value.
This is equivalent to "probability that the two first cards are aces".

First card is an ace -> 4/52
Second card is an ace -> 3/51
• May 8th 2008, 09:01 AM
ccdelia7
Yes - I understand the probabilities of both situation (thanks for clarifying). Now, however, I'm concerned with the intersection of the two probabilities. I don't know how to think of the intersection of the two constituent probabilities. Thanks.
• May 8th 2008, 09:02 AM
ccdelia7
Gotcha! I see now how to look at it! thanks!