What is the probability that the cards are of the same suit?
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.
Best Answer
Almost. You have the probability that two cards come from a particular suit. You also have to consider that there are four suits from which those cards can be selected.
Further, we typically use combination rather than permutation for this, because order of selection in a card hand is not important. (Hands are "heaps" rather than "lists"). (Though often the common factor cancels.)
We count ways to select a suit, and two cards from that suit. Divide by all ways to select two cards of any suit.
$$\mathsf P(\text{the two cards selected are of same suit}) = \frac{{^4\mathrm C_1}{^{13}\mathrm C_2}}{{^{52}\mathrm C_2}} \times\frac{2!}{2!} = \frac{{^4\mathrm C_1}{^{13}\mathrm P_2}}{{^{52}\mathrm P_2}} = \frac{4}{17} $$
There is an alternative. Whatever card is drawn 'first', there are $12$ of the $51$ cards remaining that match the 'first' card's suit. $12/51=4/17$.