1. Definitions:
Random Phenomenon:
has outcomes that are unpredictable but that nonetheless have a regular distribution in very many repetitions
 Event:
 a set of outcome(s) of a random phenomenon (a subset of the sample space)
Probability:
 of an event is the proportion of times the event occurs in many repeated trials of a random phenomenon
Independent events:
 the outcome of one trial (attempt) doesn’t influence or change the outcome of another.
Law of Large Numbers:
the long-run relative frequency of repeated independent events settles down to the true probability (theoretical probability) as the number of trials increases.

2. Definitions:
Sample Space:
 the set of all possible outcomes (use set notation)
Disjoint events (mutually exclusive):
 events that have no outcomes in common
Complement of an event:
The event not A consists of all experimental outcomes that are not in event A. This is sometimes called the complement of A and is usually denoted by Ac
Venn Diagram for mutually exclusive events A B A Ac

3. P(A  B) = P(A) + P(B) – P(A and B)
General Probability Rules:
Any probability is always between 0 & 1.
The collection of all possible outcomes has a probability of 1.
For any event A, the probability that A does not occur (the complement of A) is
P(Ac) = 1 – P(A)
General Addition rule:
 P(A  B) = P(A) + P(B) – P(A and B)   
Multiplication rule for independent events only!:
P(A  B) = P(A)*P(B)
Probability of at least one:
P(at least 1) = 1 – P(none)

4. Examples:
1. Find the sample space and determine if the outcomes are equally likely:
Roll a die
S = {1,2,3,4,5,6} Yes equally likely for a fair die
b) Toss two coins
S = {TT, TH, HT, HH} Yes equally likely for a fair coin
c) Toss four coins, the number of heads
S = {0,1,2,3,4} Not equally likely because there is only one way to get 0 heads (ex: TTTT) but there are several ways to get 1 head (ex: HTTT, THTTT, TTHT, TTTH) and several ways to get 2 and 3 heads.

5. a) A spinner is spun, P(1) = ½; P(2) = 1/3; P(3) = ¼
Examples:
2: Are the following legitimate sample spaces? Explain.
a) A spinner is spun, P(1) = ½; P(2) = 1/3; P(3) = ¼
NO because the Probability of the sample space adds to more than 1 
b) 2 coins are tossed;
P(HH) = ½; P(HT) = ¼; P(TT) = -¼ ; P(TH) = ¼
NO because probability can’t be negative

6. Examples: 3. A die is rolled, find each probably: From the sample space {1,2,3,4,5,6} we can tell that there are 3 even numbers(2,4,6) and there are 3 prime numbers(2,3,5) a) the number is even: P(even) = 3/6 = 1/2 b) the number is prime: P(prime) = 3/6 = 1/2 c) the number is even and prime: P(even and prime) = 1/6 because there is only one number that’s even and prime (2) d) the number is even or prime: P(even or prime) = 1/2 + 1/2 – 1/6 = 5/6 almost all the numbers are even or prime, except for number 1

7. 4. One card is drawn from a standard deck, find each probably: A standard deck has 52 cards, 4 suits (clubs and spades which are black, diamonds and hearts which are red), each suit has 13 cards (Ace, 2-10, Jack, Queen, King), the jack, queen and king are considered “face cards” a) the card is a king: P(king) = 4/52 = 1/13 b) the card is an ace: P(ace) = 4/52 = 1/13 c) the card is a heart: P(heart) = 13/52 = 1/4 d) the card is an ace and a king: P(ace and king) = 0 there are no cards that are both ace and king e) the card is an ace and a heart: P(ace and heart) = 1/52 there is one card that is both ace and hearts (the ace of hearts) f) the card is an ace or a king: P(ace or king) = 1/13 + 1/13 – 0 = 2/13 g) the card is an ace or a heart: P(ace or heart) = 1/13 + 1/4 – 1/52 = 4/13

8. 5. E and F are two mutually exclusive events. If P(E) =. 2 and P(F) =
5. E and F are two mutually exclusive events. If P(E) = .2 and P(F) = .1, find: a) P(E  F) – 0 = .3 b) P(E  F) 0 for mutually exclusive events the P(A  B) is always 0 6. Given that the probability of getting a blue M&M is .38, red M&M is .36, if you select 2 M&M at random, find each probability: a) Both are blue: (.38)(.38) = b) Neither is red: (.64)(0.64) = c) at least one is red: 1 – P(none are red) = 1 – (.64)(.64) =

9. 7. A bag contains 5 red marbles, 4 black marbles, and 3 white marbles
7. A bag contains 5 red marbles, 4 black marbles, and 3 white marbles. If you select 2 marbles without replacement, what is the probability of getting at least one black marble? 1 – P(none are black) 1 – (8/12)(7/11) 19/