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)

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

Re: Basic Probability Question (Deck of Cards)

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 .

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.