in a pack of 52 regular playing cards ace, jack, queen, king are numbered 1, 11, 12, 13 respectively. two cards are drawn with replacement. what is the probability that the difference of the numbers is even?
Answers
Answer:
This specific type of question deals with the use of combinations and permutations. I know that because the cards aren't being replaced after being taken out, order does matter and therefore permutations are used. So I started off with this:
(13)P(2) since at first there are 13 cards of the same suit, then 12 after one of the suit is picked out and not replaced. The total came out to be 156 ways.
For the sample space I did (52)P(2) since at first there are 52 cards to choose from, and then there are 51 cards to choose from after the first card is picked and not replaced. The total came out to be 2652 ways.
So then to find the probability of the question:
(13)P(2)/(52)P(2)= 156/2652= 1/17
What is the probability that the cards are both face cards?
I started off with this:
There are 12 face cards in total, and because cards aren't being replaced, order does matter. So since there are 12 face cards as choices in the first try, then 11 in the second I did (12)P(2) as the numerator. The total came out to be 132.
For the denominator, which is the sample space, I did (52)P(2), since you are choosing 2 out of 52 cards and got 2652 ways.
So now I do :
(12)P(2)/(52)P(2)= 132/2652= 11/221
What is the probability that the cards are a diamond and a spade?
I was a little confused on this one, but I tried. Even though I said order matters and to use permutations, I felt like for the numerator I was supposed to combinations since choosing a diamond and spade didn't matter. Since there are 13 spades and 13 diamonds, for the numerator I did (13)C(1)*(13)C(1) which equals to 169 ways of choosing a diamond and a spade.
For the sample space I stayed with (52)P(2) since order still matter on this one which equals to 2652 ways.
So the final step is :
[(13)C(1)*(13)C(1)]/(52)P(2)= 169/2652= 13/204
I would like to know if I did these questions right, because i'm a little confused between using combinations and permutations.
Thanks!
Thanks for the A2A! The correct answer has already been given (twice). I'd like to present two solutions that expand on (and explain more completely) the reasoning of the ones already given.
One is using the hypergeometric distribution, which is meant exactly for the type of problem you describe (sampling without replacement):
P(X=k)=(Kk)(N−Kn−k)(Nn)
where N is the total number of cards in the deck, K is the total number of ace cards in the deck, k is the number of ace cards you intend to select, and n is the number of cards overall that you intend to select.
The deck has 52 cards and contains 4 aces. When you draw a card, the probability of drawing an ace is 4/52=1/13
When the second card is drawn, there are only 51 cards left and since an ace has been drawn already, only 3 of them remain in the deck. Hence probability of drawing the second ace is 3/51 . The probability that both these events happen is 1/13∗3/51=0.00452 (approx). So the probability of drawing two aces is about 0.45%.
Another way of looking at this is through combinations and permutations. The number of ways in which two cards can be selected from a deck is 52C2 . This is the tot
Two cards