Consider a very simple card game, there are 16 cards. The cards have 4 ranks (Jack, King, Queen and Ace) and 4 suits (Clubs, Diamonds, Hearts and Spades). Two cards are dealt

a. What is the probability you get a Diamond card?


b. What is the probability you get two cards of the same rank?



c. What is the probability you get two cards of the same rank or a Diamond card?



d. What is the probability you don’t get a Diamond card?



e. What is the probability you get don’t get a Diamond card and you get two cards of the same rank?


Can anyone help with this? This was an example he gave at the end of class that he wants done before class today.

This post was edited by bensfriend2 on Jan 28 2016 05:57am

Very easy probability questions ... but 0fg and its 3am here

a. Wonder about the probability of the complementary event : to not get any Diamond.
Consider the first card dealt. There are 12 cards that are not Diamonds among 16.
Hence the probability the first is not a Diamond is 12/16.
Now, in this situation, there are 11 cards among the remaining 15 that are not Diamonds.
The probability of the second card dealt not being a Diamond is 11/15.

The probability of the complementary event is : (12/16) * (11/15) = 11/20
Thus, the probability of getting (at least) a Diamond is : 1 - 11/20 = 9/20

b. The first card dealt has a given rank. Among the 15 remaining cards, exactly 3 have the same rank.
So the probability of having 2 cards of the same rank is 3/15, ie : 1/5.

c. Let's call :
D = getting (at least) a Diamond
R = getting 2 cards of the same rank.

We have just seen that :
p(D) = 9/20
p(R) = 1/5

What does D ∩ R mean ? It means getting 2 cards of the same rank, with a Diamond among them.
For any 2 cards of the same rank, there are 1 chance out of 2 to have a Diamond, as you can see when checking all possibilities :
DS / DC / DH / SC / SH / CH
(D for Diamond, S for spades ...)
or by the following reasoning : the first of the 2 cards has 3 chances out of 4 not being a Diamond, the second has 2 chances out of 3 not being a Diamond, so (3/4)*(2/3) = 2/4 = 1/2 chances of not having any Diamond, and 1-1/2 = 1/2 of having a Diamond.

Hence, p(D∩R) = p(R)*1/2 = 1/10

Now, you want to know p(D∪R) :
p(D∪R) = p(D) + p(R) - p(D∩R) = 9/20 + 1/5 - 1/10 = 11/20.

d. See a.

e. See c.

This is great! thanks for much for this. I did manage to get 3 of the 5 on my own but I couldnt quite get the more difficult ones. Once I get some FG I'll throw some your way!

I'm just happy to help

Good luck.

a. there are 4 diamonds in a total of 16 cards. take the number of ways you can pick one of the diamonds nCr(4,1) and one non-diamond, nCr(12,1) and divide by the number of ways you can pick any two cards, nCr(16,2).
=> nCr(4,1)*nCr(12,1)/nCr(16,2) = 0.4 or 40%
c. the question is getting a. or b., so simply add your result from a. and b. (b. was done correctly by feanur).

This post was edited by ruedigerpk on Jan 28 2016 12:23pm

Your answer for a. stands correct if the question is to get 1 Diamond, but not 2.

You are wrong on c. , because you count twice the events when you get 2 cards of the same rank, with a Diamond among them.