# Drawing two cards from a deck of 16 (4 ranks and 4 suits)

I'm reviewing for an exam and have come across a problem marked incorrect on my homework.

There are 16 cards in a deck. The cards have 4 ranks (Jack, Queen, King, and Ace) and 4 suits (Clubs, Diamonds, Hearts, and Spades). You are dealt two cards.

What is the probability you get a Diamond card?
I misread this question when I first asnwered it, and I'm unsure how to get the correct solution. The solution page says that the solution is $\frac{9}{20}$.

What is the probability you get two cards of the same rank?
I said that once the first card is drawn, you'll have three remaining cards with that same rank out of a total of 15 cards, so you have a $\frac{3}{15} = \frac{1}{5}$ chance. This answer was the same as the solution manual. Is my logic correct?

What is the probability you don't get a Diamond card?
This is just 1 - (the solution to part a) = $\frac{11}{20}$.

What is the probability you don't get a Diamond card and you get two cards of the same rank?
Let A be the event you don't get a Diamond card.
Let B be the event you get two cards of the same rank.

$P(A' \cup B) = P(B) - P(A \cap B) = \frac{1}{5} - \frac{1}{10} = \frac{1}{10}$

I believe I have the most misunderstanding on the first question. Thank you for any assistance.

For the first question, the easiest way to calculate the probability of not pulling a diamond card, then subtracting that from 1.

The probability of not pulling a diamond card is $$\frac{12}{16}*\frac{11}{15}=\frac{132}{240}=\frac{11}{20}$$ $\frac{12}{16}$ because that's the amount of cards that don't have diamonds, and $\frac{11}{15}$ because after pulling a non-diamond card, you will have $\frac{12-1}{16-1}=\frac{11}{15}$. Then, $$1-\frac{11}{20}=\frac{9}{20}$$

The second and fourth questions you have down, and for the third question just refer back to how I solved the first question.

For the last question:

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

You can do this most easily directly. The first card must not be a diamond, but can be anything else: 12/16 or, simplified, 3/4.

There are only two possibilities for the second card: the two non-diamond cards of same rank as the first card, out of the fifteen remaining cards. 2/15.

3/4 * 2/15 = 1/10