Basic Probability Question (Deck of Cards)

Hi everyone, hoping you can all help me with this question.

A card is drawn from a shuffled deck of 52 cards, and not replaced. Then a second card is drawn. What is the probability that the second card is a king?

If possible could you please explain the process of arriving at the answer? The answer key gives me "1/13", hopefully that is correct.

Re: Basic Probability Question (Deck of Cards)

Quote:

Originally Posted by

**Pourdo** A card is drawn from a shuffled deck of 52 cards, and not replaced. Then a second card is drawn. What is the probability that the second card is a king?

Notation: $\displaystyle K_1$ is first card is a king. $\displaystyle \overline{K_1}$ is first card is **not** a king.

We want $\displaystyle \mathcal{P}(K_2)=\mathcal{P}(K_1\cap K_2)+\mathcal{P}(\overline{K_1}\cap K_2)$.

Well what is $\displaystyle \frac{4}{52}\frac{3}{51}+\frac{48}{52}\frac{4}{51} =~?$

Re: Basic Probability Question (Deck of Cards)

you may also think like that:

P( 2nd card is a K)=P( 2nd card is a Q )=P( 2nd card is a J)=P( 2nd card is a 10)=...=P( 2nd card is an A)

and surely

P( 2nd card is a K)+P( 2nd card is a Q )+P( 2nd card is a J)+P( 2nd card is a 10)+...+P( 2nd card is an A)=1

that is $\displaystyle 13 P(\text{2nd card is a K})=1$

Re: Basic Probability Question (Deck of Cards)

Hello, Pourdo!

This is a classic "trick question"

designed to complicated your thinking.

Quote:

A card is drawn from a shuffled deck of 52 cards, and not replaced.

Then a second card is drawn.

What is the probability that the second card is a King?

Let us alter the problem.

Cards are drawn one at a time from a shuffled deck without replacement.

What is the probability that the 4th card is a King?

. . $\displaystyle \begin{array}{cccccccccc}\square & \square & \square & \boxed{K} & \square & \square & \cdots \end{array}$

You are already thinking of the dozens of possibilities, aren't you?

What if the first card is a King?. What if it isn't?

What if the second card is a King?. What if it isn't?

. . . and so on.

We can disregard all of that!

Our only concern is: "Is the 4th card a King?"

Answer: .$\displaystyle P(\text{4th is King}) \:=\:\frac{4}{52} \:=\:\frac{1}{13}$

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

Further proof:

There are $\displaystyle 52!$ possible arrangements of the 52 cards.

How many of them have a King in the 4th position?

. . There are 4 choices of Kings for the 4th position.

. . The other 51 cards can be arranged in $\displaystyle 51!$ ways.

Hence, there are $\displaystyle 4\cdot51!$ ways

Therefore: .$\displaystyle P(\text{4th is King}) \:=\:\frac{4\cdot51!}{52!} \:=\:\frac{4}{52} \:=\:\frac{1}{13}$

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

In general, the probability that the $\displaystyle n^{th}$ card is a King is $\displaystyle \frac{1}{13}.$

Re: Basic Probability Question (Deck of Cards)

This is how I would think about it: I presume you know that "if an event can happen in N "equally likely ways" and M of them give event "X" then the probability of event "X" is $\displaystyle \frac{M}{N}$.

For example, there are four kings in a standard deck of 52 cards, so the probability of getting a king on the first draw is 4/52= 1/13. But you want a king on the **second** draw. Since this is "without replacement", there will be 51 cards in the deck on the second draw. But how many kings? That depends on what happened on the first draw. **If** the first draw was a king, there will be, now, 3 kings left so the probability of a king on the second draw is 3/51= 1/17. **If** the first draw was not a king, there will be, now, 4 kings left so the probability of a king on the second draw is 4/51.

We have to "weight" those two probabilities by the probability they will happen: the probability of a king on the first draw is, as before, 4/52= 1/13. The probability of any thing other than a king on the first draw is 1- 1/13= 12/13. The "weighted average" will be (1/13)(1/17)+ (12/13)(4/51)= 1/221+ 48/663= 3/661+ 48/663= 51/663= 1/13.