Probability We love Section 9.3a and b!

Most people have an intuitive sense of probability, but that intuition is often incorrect… Let’s test your intuition about probability!!!

Intuition about Probability? Find the probability of each of the following events. 1. Tossing a head on one toss of a fair coin. Two equally likely outcomes: { T, H }. Probability is 1/2. 2. Tossing two heads in a row on two tosses of a fair coin. Four equally likely outcomes: { TT, TH, HT, HH }. Probability is 1/4.

Intuition about Probability? Find the probability of each of the following events. 3. Drawing a queen from a standard deck of 52 cards. There are 52 equally likely outcomes, 4 of which are queens. Probability is 4/52, or 1/ Rolling a sum of 4 on a single roll of two fair dice. Multiplication Principle of Counting  6 x 6 = 36 equally likely outcomes. Of these, three { (1, 3), (3, 1), (2, 2) } yield a sum of 4. Probability is 3/36, or 1/12.

Intuition about Probability? Find the probability of each of the following events. 5. Guessing all 6 numbers in a state lottery that requires you to pick 6 numbers between 1 and 46, inclusive. Number of equally likely ways to choose 6 numbers from 46 numbers without regard to order?  Probability is 1/

Terminology of Probability Sample Space – the set of all possible outcomes of an experiment Event – a subset of the sample space Each sample space consists of a finite number of equally likely outcomes…

Probability of an Event (Equally Likely Outcomes) If E is an event in a finite, nonempty sample space S of equally likely outcomes, then the probability of the event E is P(E) = the number of outcomes in E the number of outcomes in S A key part of the hypothesis!!!

Rolling Dice Possible outcomes for the sum on two fair dice: { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 } So the probability of rolling a sum of 4 is 1/11, right??? NO!!!, because these sums are not equally likely…

BLUEAqua Possible outcomes when rolling a BLUE and Aqua die: Sum = 2 1-2, 2-1 Sum = 3 1-3, 2-2, 3-1 Sum = 4 1-4, 2-3, 3-2, 4-1 Sum = 5 1-5, 2-4, 3-3, 4-2, 5-1 Sum = 6 1-6, 2-5, 3-4, 4-3, 5-2, 6-1 Sum = 7 2-6, 3-5, 4-4, 5-3, 6-2 Sum = 8 3-6, 4-5, 5-4, 6-3 Sum = 9 4-6, 5-5, 6-4 Sum = , 6-5 Sum = Sum = 12 Probability Distribution OutcomeProbability 21/36 32/36 43/36 54/36 65/36 76/36 85/36 94/36 103/36 112/36 121/36 Note: The sum of the probabilities of any prob. dist. is always 1!!!

Definition: Probability Function A probability function is a function P that assigns a real number to each outcome in a sample space S subject to the following conditions: 1. for every outcome O. 2. The sum of the probabilities of all outcomes in S is The empty set…

Probability of an Event (Outcomes not Equally Likely) Let S be a finite, nonempty sample space in which every outcome has a probability assigned to it by a probability function P. If E is any event in S, the probability of the event E is the sum of the probabilities of all the outcomes in E.

Returning to our dice… Suppose that two fair dice have been rolled. State the probability of each of the following events. 1. The sum is 7.P(sum 7) = 1/6 2. The same number is rolled on both dice.P(doubles) = 1/6 3. The sum is 2 or 3.P(sum 2 or 3) = 1/12 4. The sum is a multiple of 3.P(multiple of 3) = 1/3

Strategy for Determining Probabilities 1. Determine the sample space of all possible outcomes. When possible, choose outcomes that are equally likely. 2. If the sample space has equally likely outcomes, the Probability of an event E is determined by counting: P( E ) = the number of outcomes in E the number of outcomes in S 3. If the sample space does not have equally likely outcomes, determine the probability function. (This is not always easy to do.) Check to be sure that the conditions of a probability function are satisfied. Then the probability of an event E is determined by adding up the probabilities of all the outcomes contained in E.

Multiplication Principle of Probability Suppose an event A has probability p and an event B has probability p under the assumption that A occurs. then the probability that both A and B occur is p p If the events A and B are independent, we omit the phrase “under the assumption that A occurs.”

Practice Problems Sal opens a box of a dozen chocolate covered doughnuts and offers two of them to Val. Val likes the vanilla-filled doughnuts the best, but all of the doughnuts look alike on the outside. If four of the twelve doughnuts are vanilla-filled, what is the probability that both of Val’s picks turn out to be vanilla? Here, we are choosing two doughnuts (without regard to order) from a box of 12: outcomes of this experiment Are they all equally likely?  YES!!!

Practice Problems Sal opens a box of a dozen chocolate covered doughnuts and offers two of them to Val. Val likes the vanilla-filled doughnuts the best, but all of the doughnuts look alike on the outside. If four of the twelve doughnuts are vanilla-filled, what is the probability that both of Val’s picks turn out to be vanilla? The event E consists of all possible pairs of 2 vanilla-filled doughnuts (without regard to order) from the 4 vanilla-filled doughnuts available: ways to form such pairs

Practice Problems Another way to solve the same problem:  We want the probability of outcome VV. There are two types of doughnuts: vanilla (V) and unvanilla (U) When choosing two doughnuts, there are four possible outcomes: { VV, VU, UV, UU } Are they all equally likely?  NO WAY!!!

Practice Problems There are two types of doughnuts: vanilla (V) and unvanilla (U) When choosing two doughnuts, there are four possible outcomes: { VV, VU, UV, UU } Probability of picking a V on the first draw: 4/12 Probability of picking a V on the second draw, under the assumption that a V was drawn on the first: 3/11 Probability of drawing a V on both draws (use the Mult Prop!!!):

Practice Problems Now, I want you to create a probability function for this experiment… OutcomeProbability V V1/11 V U(4/12)(8/11) = 8/33 U V(8/12)(4/11) = 8/33 U U(8/12)(7/11) = 14/33 Does this probability function “check out?”

Practice Problems Skittles candy comes in the following color proportions: Color Proportion Red 0.25 Green 0.2 Yellow 0.15 Orange 0.3 Purple 0.1 A single Skittle is selected at random from a newly-opened bag. What is the probability that the candy has the given color(s)? 1. Green or Orange P(G or O) = P(G) + P(O) = = Purple or Orange or Red P(P or O or R) = P(P) + P(O) + P(R) = = 0.65

Practice Problems Skittles candy comes in the following color proportions: Color Proportion Red 0.25 Green 0.2 Yellow 0.15 Orange 0.3 Purple 0.1 A single Skittle is selected at random from a newly-opened bag. What is the probability that the candy has the given color(s)? 3. Not Yellow P(not Y) = 1 – P(Y) = 1 – 0.15 = Neither Green nor Red P[not (G or R)] = 1 – P(G or R) = 1 – ( ) = 0.55

Practice Problems Skittles candy comes in the following color proportions: Color Proportion Red 0.25 Green 0.2 Yellow 0.15 Orange 0.3 Purple 0.1 Now, a Skittle is selected at random from each of three newly- opened bags. What is the probability that the three Skittles have the given color(s)? 5. All three are Orange P(O1 and O2 and O3) = P(O1) x P(O2) x P(O3) = (0.3)(0.3)(0.3) = 0.027

Practice Problems Skittles candy comes in the following color proportions: Color Proportion Red 0.25 Green 0.2 Yellow 0.15 Orange 0.3 Purple 0.1 Now, a Skittle is selected at random from each of three newly- opened bags. What is the probability that the three Skittles have the given color(s)? 6. All three are Purple P(P1 and P2 and P3) = P(P1) x P(P2) x P(P3) = (0.1)(0.1)(0.1) = 0.001

Practice Problems Skittles candy comes in the following color proportions: Color Proportion Red 0.25 Green 0.2 Yellow 0.15 Orange 0.3 Purple 0.1 Now, a Skittle is selected at random from each of three newly- opened bags. What is the probability that the three Skittles have the given color(s)? 7. None are Red P(none R) = P(not R1 and not R2 and not R3) = P(not R1) x P(not R2) x P(not R3) = (0.75)(0.75)(0.75) =

Practice Problems Skittles candy comes in the following color proportions: Color Proportion Red 0.25 Green 0.2 Yellow 0.15 Orange 0.3 Purple 0.1 Now, a Skittle is selected at random from each of three newly- opened bags. What is the probability that the three Skittles have the given color(s)? 8. The 1 st is Red, the 2 nd is Orange, and the 3 rd is not Purple P(R1 and O2 and not P3) = P(R1) x P(O2) x P(not P3) = (0.25)(0.3)(0.9) =