Math 140 Introductory Statistics

The Big Picture of Statistics

Lecture 10 Introducing Probability Chapter 10

Example What is probability?

Suppose that we are interested in estimating the percentage of U.S. adults who favor the death penalty. In Probability is a mathematical description of order to do so, we choose a random sample of 1200 randomness and uncertainty. U.S. adults and ask their opinion: either in favor of or Probability questions arise when we are against the death penalty. faced with a situation that involves uncertainty. Such a situation is called a We find that 744 out of the 1200, random experiment , an experiment that or 62%, are in favor. produces an outcome that cannot be predicted in advance (hence the uncertainty). "How likely is it that the percentage of US adults who favor the death penalty in our sample is off by more than 3% from the population percentage?".

The result of any single coin toss is Example: coin toss Coin toss random. But the result over many tosses is predictable, as long as the trials are independent (i.e., the outcome of a new The result of any single coin toss is random. coin toss is not influenced by the result of the previous toss).

The probability of Possible outcomes: heads is 0.5 = the proportion of times Heads (H) you get heads in many repeated trials. Tails (T) First series of tosses Second series

1 The Law of Large Numbers Sample Space

In random sampling, the larger the sample, the Each random experiment has a set of possible closer the proportion of successes in the sample tends to be to the proportion in the population. outcomes, and there is uncertainty as to which of the outcomes we are actually going to get once the experiment is conducted. This list of possible outcomes is called the sample space of the random experiment, and is denoted by the (capital) letter S.

http://bcs.whfreeman.com/ips4e/cat_010/applets/expectedvalue.html

Examples Sample space Important: It’s the question that determines the sample space.

Toss a coin once. The possible outcomes that this H - HHH A. A basketball player random experiment can produce are: {H, T}, thus the H shoots three free throws. M - HHM sample space is S = {H, T}. H What are the possible S = {HHH, HHM, H - HMH Toss a coin twice. The possible outcomes that this sequences of hits (H) and M HMH, HMM, MHH, MHM, MMH, MMM } random experiment can produce are: {HH, HT, TH, misses (M)? M - HMM M … TT}, thus the sample space is S = {HH, HT, TH, TT} … Note: 8 elements, 2 3 Chose a person at random and check his/her blood type. In this random experiment the possible B. A basketball player shoots three free throws. What is the S = {0, 1, 2, 3} outcomes are the four blood types: {A,B,AB,O}, thus number of baskets made? the sample space is S = {A,B,AB,O}. C. A nutrition researcher feeds a new diet to a young male white rat. What are the possible outcomes of weight gain (in grams)?

S = [0, ∞] = (all numbers ≥ 0)

An Event Example

Now we can talk about an event of interest, which is a statement about the nature of the outcome that we're actually going to get once the experiment is conducted. Events are denoted by capital letters (other than S, which is reserved for the sample space). Consider example 3, tossing a coin 3 times. The sample space in this case is: S = {HHH, THH, HTH, HHT, HTT, THT, TTH, TTT} We can define the following events: Event A: "Getting no H" --> TTT Event A: "Getting no H" Event B: "Getting exactly one H" --> HTT, THT, TTH Event B: "Getting exactly one H" Event C: "Getting at least one H" --> HTT, THT, TTH, Event C: "Getting at least one H" THH, HTH, HHT, HHH

2 Probability Equally Likely Outcomes

Once we define an event, we can talk about the probability of the event happening and we use the If you have a list of all possible outcomes and all notation: outcomes are equally likely, then the probability of a P(A) - the probability that event A occurs, specific outcome is P(B) - the probability that event B occurs, etc. Going back to our examples and the events we've defined: In the previous example , P(A) is the shorthand notation for "the probability of getting no 'heads' in 3 tosses of a fair coin." The probability of an event tells us how likely is it for the event to occur.

Example: roll a die Example: roll a die

Possible outcomes: Event E: getting an even number. S={1,2,3,4,5,6}

Each of these are equally likely. Event A: rolling a 2 Since 3 out of the 6 equally likely outcomes make up The probability of rolling a 2 is P(A)=1/6 the event E (the outcomes {2, 4, 6}), the probability of Event B: rolling a 5 event E is simply P(E)= 3/6 = 1/2. The probability of rolling a 5 is P(A)=1/6

Example: roll two dice A couple wants three children. What are the arrangements of boys (B) and girls (G)? What is the probability of the outcomes summing to five? Genetics tells us that the probability that a baby is a This is S: boy or a girl is the same, 0.5. {(1,1), (1,2), (1,3), ……etc.} Sample space: {BBB, BBG, BGB, GBB, GGB, GBG, BGG, GGG} All eight outcomes in the sample space are equally likely. There are 36 possible outcomes in S, all equally likely (given fair dice). Thus, The probability of each is thus 1/8. the probability of any one of them is 1/36. P(the roll of two dice sums to 5) = P(1,4) + P(2,3) + P(3,2) + P(4,1) = 4 * 1/36 = 1/9 = 0.111

3 A couple wants three children. What are the Probability numbers of girls ( X) they could have?

The same genetic laws apply. We can use the probabilities above to calculate the probability for each possible number of girls. Probability of an event ranges from 0 Sample space {0, 1, 2, 3} to 1. The closer the probability is to 0, P(X = 0) = P(BBB) = 1/8 the less likely the event is to occur. P(X = 1) = P(BBG or BGB or GBB) = P(BBG) + P(BGB) + P(GBB) = 3/8 The closer the probability is to 1, the more likely this event is to occur.

Coin Toss Example: Probability of an Event Probability rules S = {Head, Tail} Probability of heads = 0.5 Probability of tails = 0.5

1) Probabilities range from 0 Probability of getting a head = 0.5 (no chance of the event ) to We write this as: P(head) = 0.5 1 ( the event has to happen ). P(neither head nor tail) = 0 The event is more likely to 1/2 The event is more likely to 0 1 P(getting either a head or a tail) = 1 NOT occur than to occur occur than to occur For any event A, 0 ≤ P(A) ≤ 1

2) The probability of the 3) The probability of an event The event is as likely The event will The event will complete sample space must not occurring is 1 minus the to occur as it is NOT occur for CERTAIN NEVER occur equal 1. probability that does occur. to occur P(sample space) = 1 P(A) = 1 – P(not A)

P(head) + P(tail) = 0.5 + 0.5 = 1 P(tail) = 1 – P(head) = 0.5

Probability rules ( cont'd ) A and B disjoint

4) Two events A and B are disjoint if they have no outcomes in common and can never happen together . The probability that A or B occurs is the sum of their individual probabilities. P(A or B) = “ P(A U B)” = P(A) + P(B) This is the addition rule for disjoint events. A and B not disjoint

Example: If you flip two coins and the first flip does not affect the second flip, S = {HH, HT, TH, TT}. The probability of each of these events is 1/4, or 0.25.

The probability that you obtain “only heads or only tails” is: P(HH or TT) = P(HH) + P(TT) = 0.25 + 0.25 = 0.50

4 Examples Examples

Rule 1: For any event A, 0 ≤ P(A) ≤ 1 Rule 2: P(sample space) = 1

Determine which of the following numbers could represent the probability of an event? 0 1.5 -1 50% 2/3

Example Note Rule 3: P (A) = 1 – P(not A) Rule 3: P (A) = 1 – P(not A) It can be written as It can be written as P(not A) = 1 – P(A) or P(A)+P(not A) = 1 P(not A) = 1 – P(A) or P(A)+P(not A) = 1

In some cases, when finding P(A) directly is very What is probability that a randomly selected complicated, it might be much easier to find person does NOT have blood type A? P(not A) and then just subtract it from 1 to get P(not A) = 1 – P(A) = 1 – 0.42 = 0.58 the desired P(A).

Rule 4 Examples

Consider the following two events: We are now moving to rule 4 which deals with A - a randomly chosen person has blood type A, and another situation of frequent interest, finding P(A B - a randomly chosen person has blood type B. or B), the probability of one event or another Since a person can only have one type of blood flowing occurring. through his or her veins, it is impossible for the events A and B to occur together. In probability "OR" means either one or the other On the other hand...Consider the following two events: or both, and so, A - a randomly chosen person has blood type A P(A or B) = P(event A occurs or event B occurs or B - a randomly chosen person is a woman. both occur) In this case, it is possible for events A and B to occur together.

5 Disjoint or Mutually Exclusive Events Decide if the Events are Disjoint

Definition: Two events that cannot occur at the Event A : Randomly select a female worker. same time are called disjoint or mutually Event B : Randomly select a worker with a college exclusive . degree. Event A : Randomly select a male worker. Event B : Randomly select a worker employed part time. Event A : Randomly select a person between 18 and 24 years old. Event B : Randomly select a person between 25 and 34 years old.

Example We will learn more probability rules. Let’s start with the more general form of the Rule 4: P(A or B) = P(A) + P(B) addition rule. Recall: the Addition Rule for disjoint events is

P(A or B) = P(A) + P(B) What is the probability that a randomly selected person has either blood type A or B? But what rule can we use for NOT disjoint events? Since “blood type A” is disjoint of “blood type B”, P(A or B) = P(A) + P(B) = 0.42 + 0.10 = 0.52

The general addition rule The general addition rule: example

General addition rule for any two events A and B: What is the probability of randomly drawing either an ace The probability that A occurs, or a heart from a pack of 52 playing cards? or B occurs, or both events There are 4 aces in the pack and 13 hearts.

occur is: However, one card is both an ace and a heart. Thus: P(ace or heart) = P(ace) + P(heart) – P(ace and heart) ) P(A or B) = P(A) + P(B) – P(A and B = 4/52 + 13/52 - 1/52 = 16/52 ≈ 0.3

6 Note Addition Rules

The General Addition Rule works ALL the time, for ANY two events P(A or B) = P(A) + P(B) – P(A and B) Note that if A and B are disjoint events, P(A and B)=0, thus 0

P(A or B) = P(A) + P(B) – P(A and B) P(A or B)=P(A)+P(B) P(A or B)=P(A)+P(B)-P(A and B) = P(A) + P(B) since P(A and B)=0 Which is the Addition Rule for Disjoint Events.

Probability Rules Probability Rules

1. The probability P(A) for any event A is 0≤ P(A) ≤1. 1. The probability P(A) for any event A is 0≤ P(A) ≤1. 2. If S is the sample space in a probability model, then 2. If S is the sample space in a probability model, then P(S)=1. P(S)=1. 3. For any event A, P(A does not occur) = 1- P(A). 3. For any event A, P(A does not occur) = 1- P(A). 4. If A and B are disjoint events, P(A or B)=P(A)+P(B). 4. If A and B are disjoint events, P(A or B)=P(A)+P(B). 5. For any two events, 5. For any two events, P(A or B) = P(A)+P(B)-P(A and B). P(A or B) = P(A)+P(B)-P(A and B).

Probability definition Probability A correct interpretation of the statement “The probability that a From a computer simulation of rolling a fair die ten times, the child delivered in a certain hospital is a girl is 0.50” would be following data were collected on the showing face: which one of the following? What is a correct conclusion to make about the next ten rolls of the a) Over a long period of time, there will be equal proportions of same die? boys and girls born at that hospital. a) The probability of rolling a 5 is greater than the probability of b) In the next two births at that hospital, there will be exactly one rolling anything else. boy and one girl. b) Each face has exactly the same probability of being rolled. c) To make sure that a couple has two girls and two boys at that c) We will see exactly three faces showing a 1 since it is what we hospital, they only need to have four children. saw in the first experiment. d) A computer simulation of 100 births for that hospital would d) The probability of rolling a 4 is 0, and therefore we will not roll produce exactly 50 girls and 50 boys. it in the next ten rolls.

7 Random phenomena Probability models Which of the following events would NOT be If a couple has three children, let X represent the number considered a random phenomenon? of girls. Does the table below show a correct probability model for X? a) The event that the next passing car will be blue. b) The event that a student gets an answer correct after hours of studying. a) No, because there are other values that X could be. b) No, because it is not possible for X to be equal to 0. c) The event that a person’s height is bigger than their armspan. c) Yes, because all combinations of children are represented. d) The event that the next customer at a grocery store d) Yes, because all probabilities are between 0 and 1 and buys bananas. they sum to 1.

Probability Probability If a couple has three children, let X represent the If a couple has three children, let X represent the number of girls. What is the probability that the number of girls. What is the probability that the couple does NOT have girls for all three couple has either one or two girls? children?

a) 0.125 a) 0.375 b) 0.125 + 0.375 = 0.500 b) 0.375 + 0.375 = 0.750 c) 1 – 0.125 = 0.825 c) 1 - 0.125 = 0.825 d) 0.500