Question

2 cards are drawn without replacement from a standard deck of 52 cards. What is the probability that both are spades given that one is a spade.

Answer 1

So how many spades are in a deck of cards?

Answer 2

13

Answer 3

What is the probability of pulling a spade for the first card?

Answer 4

1/4

Answer 5

okay... lets keep the fractions in their original numbers... 13/52. this will make more sense.
Now, the second card, you have pulled out the one spade already... now how many spades do you have left?

Answer 6

12

Answer 7

good. If we're not going to replace that card back in to the deck... how many total cards do we have left in the deck?

Answer 8

51

Answer 9

good. so what is the probability of pulling out a spade the second time?

Answer 10

12/51

Answer 11

now.. with two card pulls, stats calls it an event... based on the word problem.. you need to pull one card and then another card... when you see the word... and it means you need to multiple the two events or ratios together.

Answer 12

hmmm

Answer 13

the question asks
"What is the probability that both are spades given that one is a spade."

Answer 14

the "given" part means you KNOW the first card was a spade
so there are 12 spades, 51 in the deck total, and the probability that you draw a spade is now \(\frac{12}{51}\) and that is the answer to the question

Answer 15

Um you don't know that the first one it the spade. The second card could also be the spade draw.

Answer 16

*is

Answer 17

right.. the probability of pulling the spades... the problem isnt asking whats the probability of not pulling a spade.. that's a different problem.

Answer 18

yes, you do know it
that is what "given" means in "2 cards are drawn without replacement from a standard deck of 52 cards. What is the probability that both are spades GIVEN that one is a spade"

Answer 19

ooh i see, it is "given one is a spade" not "given the first one is a spade"

Answer 20

Yes and also the question just before this one asked the probability given that the first draw was a spade so I don't think I would be asked the same question twice.

Answer 21

The answer is 13.3% btw. I just don't how to get that answer.

Answer 22

ok we can do this
the probability that both are spades is
\[\frac{13}{52}\times \frac{12}{51}\]
the probability that at least one of the two cards is a space is
\[1-\frac{49}{52}\times \frac{48}{51}\]

Answer 23

and so since
\[P(A|B)=\frac{P(A\cap B)}{P(B)}\] you get
\[\frac{\frac{13}{52}\times \frac{12}{51}}{1-\frac{49}{52}\times \frac{48}{51}}\]

Answer 24

there you go... after you multiple the two.. the GIVEN part of the question... is the next step

Answer 25

13% exactly which makes me wonder if there is an easier way to do it
http://www.wolframalpha.com/input/?i=%2813%2F52-12%2F51%29%2F%281-49%2F52*48%2F51%29

Answer 26

@kealiiballao yeah, i should learn to read
is there an easier way?

Answer 27

I really have no other way doing this types of problems...

Answer 28

@ArkGoLucky does that last step make sense to you?

Answer 29

1 = the probably of pulling any card
49/52 = on the first event, the probability of not pulling an ace.
48/51 = on the second event, the probably not pulling an ace the second time.

Answer 30

I understand the formula but would the probability of getting at least one spade be 1-(39/52)*(38/51)

Answer 31

spades not aces

Answer 32

oh yeah.. spades

Answer 33

Nope... because the bottom part.. is based on the opposite situation.

Answer 34

we could try this
\[\frac{\frac{13}{52}\times \frac{12}{51}}{2\times \frac{13}{52}\times \frac{49}{51}+\frac{13}{52}\times \frac{12}{52}}\]

Answer 35

numerator is both are spades
denominator is 1st is, second isn't or 1st isn't second is or both are

Answer 36

then this gives
\[\frac{13\times 12}{2\times 13\times 49+12\times 13}\]

Answer 37

hmm this isn't 13% though...

Answer 38

oh yes it is! i made a typo!!

Answer 39

i put 49 instead of 39 doh
now it is right
http://www.wolframalpha.com/input/?i=13*12%2F%282*13*39%2B13*12%29

Answer 40

did it twice!!\[\frac{\frac{13}{52}\times \frac{12}{51}}{1-\frac{39}{52}\times \frac{38}{51}}\]

Answer 41

also here \[\frac{\frac{13}{52}\times \frac{12}{51}}{2\times \frac{13}{52}\times \frac{39}{51}+\frac{13}{52}\times \frac{12}{52}}\]

Answer 42

now it is correct, you get \(13.333\%\)

Answer 43

How did you know that P(B) was first card, second card or both?

Answer 44

B is "one card is a spade" so it means one or the other or both

Answer 45

i.e. first or second or both first and second

Answer 46

Okay I think I understand.

Answer 47

let me check that both solutions give the same answer

Answer 48

yeah
they both give the same
http://www.wolframalpha.com/input/?i=%2813%2F52*12%2F51%29%2F%281-39%2F52*38%2F51%29