Chapter 4: Probability

Probability

-"the chance of an event occurring" -a number between 0 and 1, possibly including 0 and 1, that is assigned under certain fixed rules.

Set Notation

-When you have a set of numbers and you put them in "{ }" -collection of objections sometimes called elements. Elements of sets can be written inside { } -Ex. Set of possibilities from rolling a six-sided die: {1, 2, 3, 4, 5, 6} -***Empty set: set with no elements. Denoted by ø. -***Subset: set that is some number of elements of a larger set. Denoted by a capital letter like A, B, C. Inside the subset possibilities from rolling a die, let subset E be prime numbers (#s that cannot have two smaller #s multiply into it). E {2, 3, 5} -***Complement: of a subset consists of all elements of the larger set not in that subset. Denoted by Ē or E^c "E bar." Ex. E^c = {1, 4, 6}

Dice Sample Spaces: What is the sample space for one roll of a standard six-sided die? What about for the sum of two rolls of a die?

S = {1, 2, 3, 4, 5, 6} S = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

Conditional Dice

-Say we roll a red die and a blue die. -From classical probability, P(blue die = 6) = 1 / 6. -Conditioning, P(blue die = 6 | total = 12) = 1since the only possible way for total 12 is two sixes. -P(blue die = 6 | total = 9) = 1 / 4 (3, 6) (4, 5) (5, 4) (6, 3) -P(blue die = 6 | total = 7) = 1 / 6 (1, 6) (2, 5) (3, 4) (4, 3) (5, 2) (6, 1) -Sometimes, knowing information affects probability, and sometimes probability is not affected.

Mathematical Probability Rules

-The probability of any event E is a real number between and including 0 and 1. 0 ≤ P(E) ≤ 1 ② The sum of the probabilities of all the outcomes in a sample space is 1. P(S) = 1 ③ If an event cannot occur, its probability is 0. ④ If an event is certain to occur, its probability is 1. ⑤ The probability of the complement of an event is 1 minus the probability of the event. P(Ē) = 1 -P(E)because P(E) + P(Ē) = 1.

Math Note: Independence in Algebra

-Thinking back to algebra, there were independent variables X and dependent variables Y. -Independence was uni-directional. -Probability events are NOT like algebra variables! -Both events are independent of each other,or both events are dependent on each other.It's not one and one.

Counting Way Practice: -If the conference gave out medals, how many ways would there be to give gold, silver, and bronze? -How many ways are there to select a set of 2 different letters and 4 different numbers from an urn with 6 letters and 10 numbers? -In Mega Millions, a player picks 5 different numbers from 1 to 70, and a number from 1 to 25. How many possible picks are there?

-This is an ordered permutation; 10P3 = 720 -6C2 * 10C4 = 15 * 210 = 3150, using combinations & fundamental counting -70C5 * 25C1 = 302,575,350

Independent Events

-We can compare conditional probability P(A | B) with unconditional probability P(A). -If knowledge about event B does not affect the probability for event A, P(A | B) = P(A), then events A and B are independent events. -If P(A | B) ≠ P(B), with knowledge of event B leading to different conditional probability for event A, events A and B are dependent events.

Independence Multiplication Test

-We know P(A and B) = P(A)*P(B|A) = P(B)*P(A|B). -If events A and B are independent, P(B | A) = P(B) and P(A and B) = P(A) * P(B). -If events A and B are dependent, P(B | A) ≠ P(B) and P(A and B) ≠ P(A) * P(B). -This gives us another test for independence. We can check if multiplication holds, or use the old check if conditional probability equals unconditional. -Also, for mathematical reasons, the purple definition is sometimes the stated definition of independence.

Classical Probability (Cont...)

-A balanced or fair implement has equal probability assigned for each outcome. •Six sided die, deck of 52 playing cards, coin -Classical probability for an event E equals (number of outcomes in E) / (number of outcomes in S) -The classical balanced implement is an ***urn, container that contains balls of equal size and weight (i.e. lotteries & bingo games).

Probability Known and Unknown

-Examples so far have taken place without knowledge of other events. We just did them. -Sometimes this is not an issue, as knowing a result doesn't affect the probability of another event. -For instance, knowing the value of the red die doesn't change P(blue die = 1). P(blue die = 1) = 1/6. -But, if we knew that the sum of the dice was 9,we would know that P(blue die = 1) is zero.

General Addition Rule

-For any two events A and B in the sample space S, P(A or B) = P(A) + P(B) - P(A and B) -When A and B are mutually exclusive, the last term becomes zero. -For SKUNK, let A = roll 1 on first orange die, B = roll 1 on second white die, P(A) = 1/6, P(B) = 1/6 -P(A and B) = 1/36 -Thus, P(A or B) = 1/6 + 1/6 - 1/36 = 11/36 -With decimals, 0.167 + 0.167 - 0.028 = 0.306 -You can express probabilities as decimals or fractions

Intersection

-For two or more subsets inside a sample space, the intersection consists of elements in ALL subsets. -Represented by and, symbol ∩ F = {Aug, Sept, Oct, Nov, Dec} D ∩ F = {Aug, Oct, Dec}

Change is Bi-Directional

-If probability of event A is changed by knowledge of event B, then probability of event B will be changed by knowledge of event A. -We can see this in the conditional probability formula. -P(A|B) = P(A and B)/P(B) and P(B|A) = P(B and A)/P(A) P(B and A) = P(A and B) through symmetry P(A|B) * P(B) = P(A and B) = P(B|A) * P(A) by multiplication P(A|B) * P(B) = P(A) * P(B|A) -If P(A|B) ≠ P(A), then P(B|A) ≠ P(B) and if P(A|B) = P(A), then P(B|A) = P(B) since the sides of the equation must remain equal.

Frequentist Probability

-In many probability situations, we have frequency counts. -If we select objects "at random" or "randomly", we assume every object has equal likelihood of selection. -For instance, say a bag of Sour Patch Kids has 7 red, 5 green, 9 blue, 10 yellow, and 7 orange candies. -If I select a candy at random, the probability that the candy is green is 5 / 38. -What is the probability that the candy is yellow or blue? P(yellow or blue) = 10/38 + 9/38 - 0/38 = 19/38.

Mathematics, Not Causality

-In mathematical probability, the question is if probability of one event changes based on knowledge of the other event. -One event does not have to "cause" the other event to shift probability. It might just be association. -Events E "even number of days of TV" and C "3 or more days" changed probabilities, but there's no causal relationship involved. -Events "pierced" and "female" changed probabilities, and there might be causality. -Events "blue die = 6" changed given"blue die = 6 | total = 12" with definite causality. *try to use the word "change" and not causality

Math Note: Dividing Fractions

-In probability, we can have fractions within fractions, such as 124/500 / 171/500 or 1/2 / 2/3. -We could use the arithmetic rule a/b / c/d = a * d / b * c and we can drop denominators when b = d. -124/500 / 171/500 = 0.248/0.342 = 0.725 rounding to three decimal places. -Practice: Find a decimal representation of 47/247 / 343/449 Answer: 47/247 / 343/449 = 0.19028/0.76392 = 0.429

Selecting from a group

-Instead of having separate trials, more frequently we have a set of n objects, and we want to select r of these objects. -In Stat 2013, we will assume all objects are distinct, with no repeats. -We also assume objects are selected without replacement. -The formula depends on whether we care about selection order or not.

Independent Examples

-Is "pierced" independent of "female"? P(pierced) = 171 / 500 = 0.342 but P(pierced | female) = 124 / 253 = 0.490. "pierced" and "female" are dependent events. -In the TV example, are C and E independent events? P(E) = 193 / 500 = 0.386 but P(E | C) = 52/314 = 0.166. E and C are dependent events.

Math Note: Factorial

-Multiplication can get long, such as writing 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * -Instead, mathematicians often write products of counting numbers as a factorial. -The symbol for factorial is ! -For any counting number n, n factorial represents n! = n * (n-1) * (n-2) * ... * 1 -By definition, 0! = 1. -For example, 4! = 4 * 3 * 2 * 1 = 24. -The multiplication above, 10! = 3,628,800.

Conditional Probability

-When we compute probability with knowledge of another event, we compute conditional probability. -The conditional probability of an event A, given event B, is the probability that event A occurs computed with knowledge that event B happened in this trial. -Denote given information on the right side of a bar. •A|B reads as "A given B occurred" .•B|A reads as "B given A". -For example, P(wearing sorority clothes| female) likely differs from P(wearing sorority clothes| male) and P(female | wearing sorority clothes).

Probability experiment

-chance process that will lead to one out of two or more defined results called outcomes. -***Trial: process of observation or measurement. A trial may involve more than one thing--like drawing 7 cards or checking 1368 records. -***Outcome: result of a single trial. -***Event: subset of one or more outcomes inside a sample space

Union

-for two or more subsets inside a sample space, the union consists of elements in one or more subsets. -Represented by inclusive or, symbol ∪ -Let F be the set of months at least partially in the Fall 2018 semester. F = {Aug, Sept, Oct, Nov, Dec} D "31 day months" ∪ F = {Jan, Mar, May, July, Aug, Sept, Oct, Nov, Dec}

Fundamental Counting Principle

-if an event can happen in N ways, and another, independent event can happen in M ways, then both events together can happen in N x M ways. -Uses multiplication of the number of ways each event in an experiment can occur to find the number of possible outcomes in a sample space. -If there are n separate trials, where the first trial has k1 possible outcomes, second has k2 possibilities, ..., and nth trial has kn possibilities, the total number of possible outcomes is k1 * k2 * .... * kn -If I want to roll a die, and separately pick one of 8 crayon colors, there are 6 * 8 = 48 possible outcomes. -If I want to select one football team from •The Big 12, with 10 teams •The Big Ten, with 14 teams •The Pac-12, with 12 teams •The SEC, with 14 teams •The ACC, with 14 teams -There are 10 * 14 * 12 * 14 * 14 = 329,280 possible outcomes.

Mutually exclusive "disjoint"

-two events that have no elements/outcomes in common and so can never occur together. -For mutually exclusive sets A and B, A ∩ B = ø Are these subsets mutually exclusive? "Subset of odd numbers and subset of even numbers." YES b/c in one roll/trial we can get an odd and even number on a single die "Subset of prime numbers and subset of even numbers." NO b/c we could get 2, 2 is both prime & even Let S be the set of directions labeled on a highway. Are the subsets "labeled east" and "labeled west" mutually exclusive? NO, near OKC airport, the road is labeled West I-44 and East OK-3. *P(A or B) = P(A)+P(B) - P(A and B) // Let C = sum of two dice is 9 or higher P(C) = 10/36 P(A and C) = 0 b/c cannot happen with six-sided die "mutually exclusive" P(A or C) = 1/6 + 10/36 - 0 = 16/36

Classical probability

-uses sample spaces to determine the numerical probability that an event will happen. -assumes that all outcomes in the sample space [the set of all possible outcomes of a probability experiment--denoted by S] are equally likely to occur. -Ex. die (1/6) & cards (1/52) Formula: P(E) = # of outcomes in E / total # of outcomes in sample space = n(E) / n(S)

What are the three mathematical rules on counting rules that are called combinatorics?

1. Fundamental counting rule 2. Permutation rule 3. Combination rule

What are the three major interpretations for probability measurements?

1. Theoretical: computed through mathematical definitions, rules, and assignments (i.e. rolling a six-sided die) 2. Empirical: Frequency portion of the time that events of the same type will occur in the long run. (i.e. how many jellybeans are seen in a jar) *relies on actual experience to determine the likelihood of outcomes. 3. Subjective: Assigned estimate of chance, considering data, experience, and personal belief. (i.e. asking someone out on a date or sports betting) *psychological evidence favors subjectivity

Losing in SKUNK: On a single roll, what is the probability of getting a 1 and ending the round?

11/36: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),(2, 1), (3, 1), (4, 1), (5, 1), (6, 1) Don't count (1, 1) twice!

Sample space practice: "Possible Birth Months" Let D be the set of months with 31 days. D = {Jan, Mar, May, Jul, Aug, Oct, Dec} What is complement D^c? Sample space is flipping a coin... S = {heads, tails}

D^c = {Feb, Apr, Jun, Sept, Nov} *In this class, we will ignore degenerate outcomes like "side."

The Dreaded Word Problem Example: In a group of 35 children, 14 have blond hair. Of the blond children, 8 have blue eyes. Of the non-blond children, 5 have blue eyes. Find P(eyes are not blue). Find P(blond hair or blue eyes).

Find P(eyes are not blue) = add blue eyes (13/35), subtract from total = 22/35 Find P(blond hair or blue eyes) = 14/35 + 13/35 - 8/35 = 19/35

Conditional Probability Formula

P(B|A) = P(A and B) / P(B) -Vertical bar always is right side given -Joint "and" probability is numerator. -Given event probability is denominator. -This applied in our earlier examples. -P(blue die = 6 | total = 9) = 1/36 / 4/36 = 1/4 -P(female | pierced) = P(female and Pierced)/P(Pierced) = 124/500 / 171/500 = 124/171

General Addition Practice: Events C and D are in the same sample space. Let P(C) = 0.19, P(D) = 0.37, P(C and D) = 0.12. Find P(C or D). If we draw a card from a standard 52-card deck, what is the probability it is a heart or a Queen?

P(C or D) = 0.19 + 0.37 - 0.12 = 0.44 P(heart) = 13/52 P(Queen) = 4/52 P(heart ∩ Queen) = 1/52 Thus, P(heart U Queen) = 16/52.

Watching TV Example: A survey asked 500 18-34 year old people how many days out of the last 5 they watched a TV (not computer or phone video). -106 said they watched no TV. -45 said they watched TV on (exactly) one of the last 5 days. -35 said two days. -27 said three days. -52 said four days. -235 said all five days. Define events... C: person watched TV at least 3 out of 5 days E: person watched TV an even number of days Assume a person is selected at random. Find P(C). Find P(E). Find P(C or E).

P(C) = 27/500 + 52/500 + 235/500 = 314/500 P(E) = 106/500 + 35/500 + 52/500 = 193/500 P(C and E) = 52/500 Using general addition rule, P(C or E) = 314/500 + 193/500 - 52/500 = 455/500 Also could use complement rule. Since only 1 day is not in C or E, P(C or E) = 1 - 45/500 = 455/500

Body Piercings Example: Pierced Not Pierced Total Male 47 200 247 Female 124 129 253 Total 171 329 500 Find.... P(pierced) = P(male^c) = P(pierced or female) =

P(pierced) = 171/500 P(male^c) = 1 - 247/500 = 253/500 P(pierced or female) = 171/500 + 253/500 - 124/500 = 300/500 *Conditional Example: P(pierced) = 171/ 500 P(pierced | male) = 47 / 247 P(pierced | female) = 124 / 253 P(female | pierced) = 124 / 171

Classical Probability Examples: What is the probability of rolling a 1 on a six-sided die? What is the probability of rolling an even number? An urn contains 6 lettered and 10 numbered balls. What is the probability of drawing a lettered ball?

P(roll 1) = 1/6 P(even) = 3/6 = P(roll 2) + P(roll 4) + P(roll 6) = 1/6+1/6+1/6 P(letter) = 6/16 = 3/8 = 0.375

Urns Results Example: Say we randomly draw a ball from the urn with 10 numbers and 6 letters, look at the ball then put it back, and then randomly draw a second ball. The two balls could be the same (1/16 probability). Define event A = first ball was a letter and event B = second ball was a letter Find P(A or B). Are A and B independent events?

Urn has 10 numbers and 6 letters. A = first ball was letter, B = second ball was letter -P(A or B) = P(A) + P(B) - P(A and B) P(A) = 6/16, P(B) = 6/16, P(A and B) = 6/16 * 6/16 so P(A or B) = 6/16 + 6/16 - 36/256 = 156/256 = 0.609 -Yes, since P(A|B) = P(A and B) / P(B) = 36/256 / 6/16 = 6/16 = P(A).

Tree diagram

We could use a tree diagram, adding in probabilities. _Used to determine all outcomes of a probability experiment. -They are conditional, so probabilities multiply, like P(number, letter) = P(first draw number) * P(second draw letter | first draw number)

Combination Formula & Example "Choosing a Group:" If we want to choose r objects, without regard of order, from a set of n objects, we have a combination. There are fewer combinations than permutations because order doesn't matter. •First favorite Blossom, second favorite Gandalf •First favorite Gandalf, second favorite Blossom How many ways are there to choose a group of 2 plush dolls from the set of 5?

nCr = (n r) = n! / r!(n-r)! -Bluman's book uses nCr but many other sources use two stacked numbers inside parentheses.Combinations are often called "choose", like "5 choose 2". To choose a group of 2 plush dolls out of 5, 5C2 = 5!/2!(5-2)! = 5!/2!3! = 5*4*3*2*1/2*1*3*2*1 = 10

Permutation Formula & Example "Choosing in Order:" Say we have 5 plush dolls: Bubbles (blue), Blossom (red), Buttercup (green), Mojo Jojo, and Gandalf How many ways are there to choose a favorite? How many ways are there to choose a favorite and a second favorite? If we want to choose r objects in a specific order from a set of n objects, we have a permutation.

nPr = n!/(n-r)! For the 1st favorite, 5P1 = 5!/(5-1)! = 5!/4! = 5*4*3*2*1/4*3*2*1 = 5 possible favorites. For 1st favorite and 2nd favorite, 5P2 = 5!/(5-2)! = 20