Introducing Probability

The Big Picture of Statistics Math 140 Introductory Statistics Lecture 10 Introducing Probability Chapter 10 2 Example
What is probability? Suppose that we are interested in estimating the percentage of U.S. adults who favor the death penalty. In order to do so, we choose a random sample of 1200 U.S. adults and ask their opinion: either in favor of or against the death penalty. We find that 744 out of the 1200, or 62%, are in favor. Probability is a mathematical description of randomness and uncertainty.
Probability questions arise when we are faced with a situation that involves uncertainty. Such a situation is called a random experiment, an experiment that 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?". Example: coin toss
Coin toss
The result of any single coin toss is random. But the result over many tosses is predictable, as long as the trials are The result of any single coin toss is random. independent (i.e., the outcome of a new coin toss is not influenced by the result of the previous toss).
The probability of heads is 0.5 = the proportion of times you get heads in many repeated trials. Possible outcomes:
Heads (H)
Tails (T) First series of tosses Second series 1 The Law of Large Numbers
In random sampling, the larger the sample, the closer the proportion of successes in the sample tends to be to the proportion in the population. Sample Space

Each random experiment has a set of possible 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:


Toss a coin once. The possible outcomes that this random experiment can produce are: {H, T}, thus the sample space is S = {H, T}. Toss a coin twice. The possible outcomes that this random experiment can produce are: {HH, HT, TH, TT}, thus the sample space is S = {HH, HT, TH, TT} Chose a person at random and check his/her blood type. In this random experiment the possible outcomes are the four blood types: {A,B,AB,O}, thus the sample space is S = {A,B,AB,O}.
A. A basketball player shoots three free throws. What are the possible sequences of hits (H) and misses (M)?


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" Event B: "Getting exactly one H" Event C: "Getting at least one H"
H
H - HHH
M - HHM
H - HMH
M -
H
M
M … HMM
…
B. A basketball player shoots three free throws. What is the number of baskets made?

S = {HHH, HHM, HMH, HMM, MHH, MHM, MMH, MMM }
Note: 8 elements, 23 S = {0, 1, 2, 3}
C. A nutrition researcher feeds a new diet to a young male white rat. What are the possible outcomes of weight gain (in grams)?
An Event It’s the question that determines the sample space. S = [0, ∞] = (all numbers ≥ 0) Example


Event A: "Getting no H" --> TTT Event B: "Getting exactly one H" --> HTT, THT, TTH Event C: "Getting at least one H" --> HTT, THT, TTH, THH, HTH, HHT, HHH 2 Probability





Once we define an event, we can talk about the probability of the event happening and we use the notation: P(A) - the probability that event A occurs, 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. Equally Likely Outcomes
If you have a list of all possible outcomes and all outcomes are equally likely, then the probability of a specific outcome is Example: roll a die Example: roll a die Possible outcomes: S={1,2,3,4,5,6}
Event E: getting an even number.
Since 3 out of the 6 equally likely outcomes make up the event E (the outcomes {2, 4, 6}), the probability of event E is simply P(E)= 3/6 = 1/2.
Each of these are equally likely.
Event A: rolling a 2 The probability of rolling a 2 is P(A)=1/6
Event B: rolling a 5 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? This is S: {(1,1), (1,2), (1,3), ……etc.}
There are 36 possible outcomes in S, all equally likely (given fair dice). Thus, Genetics tells us that the probability that a baby is a boy or a girl is the same, 0.5.
Sample space: {BBB, BBG, BGB, GBB, GGB, GBG, BGG, GGG}
All eight outcomes in the sample space are equally likely.
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 numbers of girls (X) they could have? Probability The same genetic laws apply. We can use the probabilities above to calculate the probability for each possible number of girls.
Probability Sample space {0, 1, 2, 3} P(X = 0) = P(BBB) = 1/8
P(X = 1) = P(BBG or BGB or GBB) = P(BBG) + P(BGB) + P(GBB) = 3/8
Probability of an Event of an event ranges from 0 to 1. The closer the probability is to 0, the less likely the event is to occur. The closer the probability is to 1, the more likely this event is to occur.
1) 0 The event is more likely to 1/2 NOT occur than to occur The event will NEVER occur The event is more likely to occur than to occur The event is as likely to occur as it is NOT to occur
Coin Toss Example: = {Head, Tail}
Probability of heads = 0.5
Probability of tails = 0.5 Probability rules 1 The event will occur for CERTAIN Probability rules (cont'd) Probabilities range from 0 (no chance of the event) to 1 (the event has to happen).
For any event A, 0 ≤ P(A) ≤ 1
S
Probability of getting a head = 0.5 write this as: P(head) = 0.5
P(neither head nor tail) = 0
P(getting either a head or a tail) = 1
We 2) The probability of the complete sample space must equal 1. P(sample space) = 1 3) The probability of an event not occurring is 1 minus the probability that does occur. P(A) = 1 – P(not A) P(head) + P(tail) = 0.5 + 0.5 = 1 P(tail) = 1 – P(head) = 0.5 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 P(not A) = 1 – P(A)
or P(A)+P(not A) = 1
What is probability that a randomly selected person does NOT have blood type A? P(not A) = 1 – P(A) = 1 – 0.42 = 0.58 Rule 4
We are now moving to rule 4 which deals with another situation of frequent interest, finding P(A or B), the probability of one event or another occurring. In probability "OR" means either one or the other or both, and so, P(A or B) = P(event A occurs or event B occurs or both occur) or P(A)+P(not A) = 1 In some cases, when finding P(A) directly is very complicated, it might be much easier to find P(not A) and then just subtract it from 1 to get the desired P(A). Examples

It can be written as P(not A) = 1 – P(A)
Consider the following two events:  A - a randomly chosen person has blood type A, and  B - a randomly chosen person has blood type B. Since a person can only have one type of blood flowing through his or her veins, it is impossible for the events A and B to occur together. On the other hand...Consider the following two events:  A - a randomly chosen person has blood type A  B - a randomly chosen person is a woman. In this case, it is possible for events A and B to occur together. 5 Disjoint or Mutually Exclusive Events Definition: Two events that cannot occur at the same time are called disjoint or mutually exclusive. Decide if the Events are Disjoint


Event A: Randomly select a female worker. Event B: Randomly select a worker with a college 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

Rule 4: P(A or B) = P(A) + P(B)

We will learn more probability rules. Let’s start with the more general form of the 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? Since “blood type A” is disjoint of “blood type B”, P(A or B) = P(A) + P(B) = 0.42 + 0.10 = 0.52 But what rule can we use for NOT disjoint events? 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) + P(B) Which is the Addition Rule for Disjoint Events.
P(A or B)=P(A)+P(B) since P(A and B)=0 Probability Rules The probability P(A) for any event A is 0≤ P(A) ≤1. If S is the sample space in a probability model, then P(S)=1. For any event A, P(A does not occur) = 1- P(A). If A and B are disjoint events, P(A or B)=P(A)+P(B). For any two events, P(A or B) = P(A)+P(B)-P(A and B). 1. 2. 3. 4. 5. P(A or B)=P(A)+P(B)-P(A and B) Probability Rules 1. 2. 3. 4. 5. The probability P(A) for any event A is 0≤ P(A) ≤1. If S is the sample space in a probability model, then P(S)=1. For any event A, P(A does not occur) = 1- P(A). If A and B are disjoint events, P(A or B)=P(A)+P(B). For any two events, 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 child delivered in a certain hospital is a girl is 0.50” would be which one of the following? From a computer simulation of rolling a fair die ten times, the following data were collected on the showing face: a) b) c) d) Over a long period of time, there will be equal proportions of boys and girls born at that hospital. In the next two births at that hospital, there will be exactly one boy and one girl. To make sure that a couple has two girls and two boys at that hospital, they only need to have four children. A computer simulation of 100 births for that hospital would produce exactly 50 girls and 50 boys. What is a correct conclusion to make about the next ten rolls of the same die? a) The probability of rolling a 5 is greater than the probability of rolling anything else. b) Each face has exactly the same probability of being rolled. c) We will see exactly three faces showing a 1 since it is what we saw in the first experiment. d) The probability of rolling a 4 is 0, and therefore we will not roll it in the next ten rolls. 7 Random phenomena Probability models Which of the following events would NOT be considered a random phenomenon? If a couple has three children, let X represent the number of girls. Does the table below show a correct probability model for X? a) b) c) d) The event that the next passing car will be blue. The event that a student gets an answer correct after hours of studying. The event that a person’s height is bigger than their armspan. The event that the next customer at a grocery store buys bananas. a) b) c) d) No, because there are other values that X could be. No, because it is not possible for X to be equal to 0. Yes, because all combinations of children are represented. Yes, because all probabilities are between 0 and 1 and they sum to 1. Probability Probability If a couple has three children, let X represent the number of girls. What is the probability that the couple does NOT have girls for all three children? If a couple has three children, let X represent the number of girls. What is the probability that the couple has either one or two girls? a) b) c) 0.125 0.125 + 0.375 = 0.500 1 – 0.125 = 0.825 a) b) c) d) 0.375 0.375 + 0.375 = 0.750 1 - 0.125 = 0.825 0.500 8