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.

Thank you in advance :).

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- Jan 22nd 2010, 04:27 PMflywithmeDrawing 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.

Thank you in advance :). - Jan 22nd 2010, 04:54 PMSoroban
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?

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

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

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 tocard: "Is it a King?"*any*

The probability will be: .$\displaystyle \frac{4}{52} \,=\,\frac{1}{13}$

Get it?

- Jan 22nd 2010, 04:57 PMPlato
- Jan 22nd 2010, 06:03 PMflywithme
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. - Jan 22nd 2010, 10:12 PMmatheagle
I always talk about this in class.

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)

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

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

$\displaystyle = {4\over 52}= {1\over 13} $ - Jan 22nd 2010, 10:17 PMVonNemo19