small easy problem

**Chiclette** · June 22nd 2008, 03:48 AM

You have a deck of cards. You random take 2 cards from it (without putting the first one back). Now i need to know what's the chance that one of those 2 cards is a spade.
maybe very easy, but some people doubted my answers and I would like to know the thruth

**Moo** · June 22nd 2008, 04:26 AM

Hi

There are 13 spades in a deck of 52 cards.

Therefore, the probability of picking a spade the first time is $\frac{13}{52}$ .

Then, there remain 12 spades in a deck of 51 cards.
So the probability that the second card you pick is a spade is $\frac{12}{51}$

Thus, the probability of picking two spades is :

$\mathcal{P}=\frac{13}{52} \cdot \frac{12}{51}=\frac 14 \cdot \frac{4}{17}=\frac{1}{17}$

I hope I'm not wrong

**angel.white** · June 22nd 2008, 04:55 AM

Originally Posted by Moo

Hi

There are 13 spades in a deck of 52 cards.

Therefore, the probability of picking a spade the first time is $\frac{13}{52}$ .

Then, there remain 12 spades in a deck of 51 cards.
So the probability that the second card you pick is a spade is $\frac{12}{51}$

Thus, the probability of picking two spades is :

$\mathcal{P}=\frac{13}{52} \cdot \frac{12}{51}=\frac 14 \cdot \frac{4}{17}=\frac{1}{17}$

I hope I'm not wrong

I don't know probability yet, taking that next semester, but what if you pull a not spade the on the first card?

Following your lovely example I get:
$\frac {39}{52} * \frac {13}{51} = \frac {13}{68}$

**Moo** · June 22nd 2008, 04:59 AM

Originally Posted by angel.white

I don't know probability yet, taking that next semester, but what if you pull a not spade the on the first card?

You're supposing you have taken one, that's why you multiply the probabilities.

---------------------------------
Further explanation :
Let's denote the event A : I pick a spade the first time.
Let's denote the event B : I pick a spade the second time.

We want the probability of $\{A \text{ and } B\}$ :

$\mathcal{P}(A \cap B)=\mathcal{P}(B/A)\mathcal{P}(A)$ .
This is the formula of conditional probability (Conditional probability - Wikipedia, the free encyclopedia).

$\mathcal{P}(B/A)$ is the probability that : "B happens when A has happened". This is why you have to consider that you pick a spade the first time

**CaptainBlack** · June 22nd 2008, 05:03 AM

Originally Posted by angel.white

I don't know probability yet, taking that next semester, but what if you pull a not spade the on the first card?

Following your lovely example I get:
$\frac {13}{52} * \frac {13}{51} = \frac {13}{204}$

Not a spade has probability $39/52$ on the first draw. Then:

$\frac{39}{52}\times \frac{13}{51}$

is the probability of drawing anything but a spade on the first card and a spade on the second.

RonL

**Moo** · June 22nd 2008, 05:04 AM

Originally Posted by angel.white

Following your lovely example I get:
$\frac {13}{52} * \frac {13}{51} = \frac {13}{204}$

This would be for example, the probability of getting a heart on the first picking, and a spade on the second picking

$\frac{13}{52}$ : probability to get one type of cards (heart, spade, etc.. i don't know the other names

)

$\frac{13}{51}$ : probability of getting one type of cards, where there are 13 cards of this type, among the remaining 51 cards...

I don't know if it's clear enough ?

**angel.white** · June 22nd 2008, 05:25 AM

Originally Posted by CaptainBlack

Not a spade has probability $39/52$ on the first draw. Then:

$\frac{39}{52}\times \frac{13}{51}$

is the probability of drawing anything but a spade on the first card and a spade on the second.

RonL

Fixed it, thank you.

Originally Posted by Moo

This would be for example, the probability of getting a heart on the first picking, and a spade on the second picking

$\frac{13}{52}$ : probability to get one type of cards (heart, spade, etc.. i don't know the other names

)

$\frac{13}{51}$ : probability of getting one type of cards, where there are 13 cards of this type, among the remaining 51 cards...

I don't know if it's clear enough ?

Since the question is "what's the chance that one of those 2 cards is a spade."

It seems to me that the situation of a ~spade on the first card is a valid scenario.

If we interpret the question to mean exactly one card is a spade then we would have to calculate getting:
S, ~S and ~S, S

Or if we interpret it to mean at least one card is a spade, then we would have to calculate:
S, S and S, ~S and ~S, S
(which I suspect is 100% minus probability of ~S,~S)

If this is wrong than I am misunderstanding the question.

**Moo** · June 22nd 2008, 05:29 AM

Originally Posted by angel.white

If this is wrong than I am misunderstanding the question.

That's because I misread the question

Ok, let A be the number of spades that are picked (A is 0,1 or 2).

We want $P(A \ge 1)$ , that is to say $1-P(A<1)=1-P(A=0)$ .

So it's $1-\frac{39}{52} \cdot \frac{38}{51}=\frac{15}{34}$

**angel.white** · June 22nd 2008, 05:38 AM

Originally Posted by Moo

$P(A \ge 1)$ , that is to say $1-P(A<1)=1-P(A=0)$ .

Hey, I got that right ^_^ Guess some of that set theory in discrete math stuck afterall :P

**Chiclette** · June 22nd 2008, 07:37 AM

thx for the help, it was indeed 1 spade or more

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