# Drawing cards without replacement.

• January 22nd 2010, 05:27 PM
flywithme
Drawing cards without replacement.
Two cards are drawn without replacement from a shuffled deck of 52 cards. What is the probability that the second card is a king?
Solutions showing step-by-step is very much appreciated.
• January 22nd 2010, 05:54 PM
Soroban
Hello, flywithme!

Quote:

Two cards are drawn without replacement from a shuffled deck of 52 cards.
What is the probability that the second card is a King?

I can solve this The Long Way, but why bother?

$P(\text{2nd is King}) \;=\;\frac{4}{52} \:=\:\frac{1}{13}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Do we care what the first card is? . . . No!

Suppose they asked for the probability that the 17th card is a King.
Do we need to know if any of the first 16 cards are Kings? . . . No.

Spread the cards face down on the table.
. . Point to any card: "Is it a King?"
The probability will be: . $\frac{4}{52} \,=\,\frac{1}{13}$

Get it?

• January 22nd 2010, 05:57 PM
Plato
Quote:

Originally Posted by flywithme
Two cards are drawn without replacement from a shuffled deck of 52 cards. What is the probability that the second card is a king? Solutions showing step-by-step is very much appreciated.

Why don't you show us some effort on your part?

EDIT: There is always a spoil-sport who must show that he can do the question.
• January 22nd 2010, 07:03 PM
flywithme
Thank you very much Soroban, I never tackled probability with abstract thoughts.
@Plato, how can I show effort on my part? I struggle hard with probability and I can't help others with other types of math because I am only mediocre.
• January 22nd 2010, 11:12 PM
matheagle
I usually ask what is the probability we select the ace of spades on the second, fifth... pick?
It's always 1/52.
BUT then I prove it....

P(King on second pick)=P(KING, KING)+P(not a KING, KING)

$= \left({4\over 52}\right)\left({3\over 51}\right) +\left({48\over 52}\right)\left({4\over 51}\right)$

$= \left({4\over 52}\right)\left[{3\over 51}+{48\over 51}\right]$

$= {4\over 52}= {1\over 13}$
• January 22nd 2010, 11:17 PM
VonNemo19
Quote:

Originally Posted by Plato
EDIT: There is always a spoil-sport who must show that he can do the question.

I'm sorry, but I really can't help but laugh... (Rofl)