Question

Four cards are randomly chosen (without replacement) from a deck of 52 cards. Let K be the event that two of the four cards are King and the other two are Aces. Let E be the event that one of the four cards is a King of Hearts. The conditional probability P(K | E) is equal to

Answer 1

is the answer 0.01531? thanks!

Answer 2

else, do let me know the steps leading to the solution! thanks!!

Answer 3

The events K and E are independent, the reason being that the probability of K occurring is unaffected by whether or not E has occurred. In this case\[P(K|E)=P(K)\]
P(K) can be found using the hypergeometric distribution.

Answer 4

thanks! So, will this be correct then:

P(K)=4!⋅4^2⋅3^2/(52⋅51⋅50⋅49⋅4) ?

Answer 5

First I calculated the probability of getting 2 kings in a sample of 4 cards without replacement:\[P(2\ kings)=\frac{\left(\begin{matrix}4 \\ 2\end{matrix}\right)\left(\begin{matrix}48 \\ 2\end{matrix}\right)}{\left(\begin{matrix}52 \\ 4\end{matrix}\right)}=0.0249995\]
The probability of getting 2 aces in a sample of 4 cards without replacement will be the same as P(2 kings).
The events 2 kings and 2 aces are independent. therefore the probability of getting 2 kings and 2 aces is0.0249995^2 = 0.000625

Answer 6

@kropot72 didn't you forget to take into account that one of the kings has to be a heart?

Answer 7

I think they're also dependent events aren't they... we're finding the probability that we get 2 kings and 2 aces, given that one of the cards is the king of hearts.

Answer 8

@agent0smith The question states that four cards are randomly chosen (without replacement). If you start looking at the selection and notice one of the cards is a King of Hearts does that affect the probability of two Kings and two Aces? I think not. That was the view taken in my attempt at a solution.
However on reconsideration I accept your point in which case we get
\[P(K|E)=\frac{P(K \cap E)}{P(E)}\]

Answer 9

But it's the probability that we get 2 K and 2 A, *given* one card is already a K of hearts. That should definitely affect the probability.

Answer 10

ie we can't have a K of spades and a K of clubs, or K diamonds and K clubs etc. Having one heart limits our possible outcomes, and thus makes K depend on E.

Answer 11

I am not sure whether you agree or disagree with the conditional probability of K given E being expressed as\[P(K|E)=\frac{(K \cap E)}{P(E)}\]
Note that the numerator on the right hand side is K intersection E which is the redefined sample space resulting from the consideration of the condition that a King of Hearts is included.

Answer 12

No I wasn't disagreeing with that, I think that's right. I was just describing why the events are dependent.

Answer 13

Sure. I thought I covered that in saying previously that I accept your point :)

Answer 14

hehe, I just wasn't sure what you meant above, if you agreed or not :P