Probability (Multiple independent events: AND)

Calculate the probability of the following sets of events:

You have cupboard full of plates, but you do not have a full set of any dishes. You have 3 blue plates, 6 flowered plates, 2 white plates and 1 green plate.

1. If you are eating alone, and you choose randomly, what is the probability that you get the green plate? p(green)=$\dfrac{1}{12}$
2. If you are eating with one other person, what's the probability that you get the green plate and a blue plate? p(green) x p(blue)=$\dfrac{1}{12} \times \dfrac{3}{11}=\dfrac{3}{132}=\dfrac{1}{44}$
3. Or, if you prefer matching plates, what's the probability that you pull out two blue plates? p(blue) x p(blue)=$\dfrac{3}{12} \times \dfrac{2}{11}=\dfrac{6}{132}=\dfrac{1}{22}$
4. If you choose randomly, what is the probability that you will pull out 4 flowered plates in a row? p(flower) x p(flower) x p(flower) x p(flower)=$\dfrac{6}{12}\times \dfrac{5}{11}\times \dfrac{4}{10}\times \dfrac{3}{9}=\dfrac{1}{33}$
5. Let's say you hate the flowered plates, what's the probability that you pull 2 blue plates and 2 white plates?p(white) x p(white) x p(blue) x p(blue)=$\dfrac{2}{12}\times \dfrac{1}{11}\times \dfrac{3}{10}\times \dfrac{2}{9}=\dfrac{12}{11880}=\dfrac{1}{990}$
6. What's the probability of getting 4 non-flowered plates? p(not flower) x p(not flower) x p(not flower) x p(not flower)=$\dfrac{6}{12}\times \dfrac{5}{11}\times \dfrac{4}{10}\times \dfrac{3}{9}= \dfrac{1}{33}$

You are playing cards with a friend.  There are 52 cards in a standard deck, divided into 4 suits (13 hearts, 13 spades, 13 clubs, 13 diamonds).  Each suit contains an ace, king, queen, jack, 10, 9, 8, 7, 6, 5, 4, 3, and 2 (there are 4 of each of these cards in each deck).  (To simplify these probabilities, we've specified the order in which the cards should be received, or limited the number of cards in a hand -- even if those are are the specific rules of the card games).

1. What is the probability that you get a pair of aces in your first two cards? p(ace) x p(ace)=$\dfrac{4}{52} \times \dfrac{3}{51}=\dfrac{12}{2652}=\dfrac{1}{221}$
2. What is the probability that you get 3 kings in the first three cards? p(king) x p(king) x p(king)=$\dfrac{4}{52} \times \dfrac{3}{51} \times \dfrac{2}{50}=\dfrac{1}{5525}$
3. What is the probability that you get a flush, all 5 of the cards in your hand are the same suit?  p(any card) x p(same suit) x p(same suit) x p(same suit) x p(same suit)=$1 \times \dfrac{12}{51} \times \dfrac{11}{50} \times \dfrac{10}{49} \times \dfrac{9}{48}= \dfrac{11880}{5997600}= \dfrac{1188}{599760}=\dfrac{99}{49980}=\dfrac{33}{16660}$
4. What is the probability that you get a royal straight (specifically, get the following cards in this order: 10, Jack, Queen, King, Ace)? p(ten) x p(jack) x p(queen) x p(king) x p(ace)=$\dfrac{4}{52} \times\dfrac{4}{51} \times \dfrac{4}{50} \times \dfrac{4}{49} \times \dfrac{4}{48}= \dfrac{1024}{311875200}= \dfrac{16}{4873050}=\dfrac{8}{2436525}$
5. What is the probability that you get a full house (specifically, three of a kind and then a pair)? p(any) x p(same card) x p(same card) x p(any card) x p(same card)=$1 \times\dfrac{3}{51} \times \dfrac{2}{50} \times 1 \times \dfrac{3}{48}= \dfrac{18}{122400}= \dfrac{1}{6800}$
6. What is the probability that you get four kings and an ace (in that order)? p(king) x p(king) x p(king) x p(king) x p(ace)=$\dfrac{4}{52}\times\dfrac{3}{51} \times \dfrac{2}{50} \times \dfrac{1}{49} \times \dfrac{4}{48}= \dfrac{96}{311875200}= \dfrac{1}{3248700}$