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# Probability - PowerPoint PPT Presentation

Probability. Assigning Probabilities. A probability is a value between 0 and 1 and is written either as a fraction or as a decimal fraction. A probability simply is a number between 0 and 1 that is assigned to a possible outcome of a random circumstance.

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## PowerPoint Slideshow about 'Probability' - denise

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### Probability

• A probability is a value between 0 and 1 and is written either as a fraction or as a decimal fraction.

• A probability simply is a number between 0 and 1 that is assigned to a possible outcome of a random circumstance.

• For the complete set of distinct possible outcomes of a random circumstance, the total of the assigned probabilities must equal 1.

Assigning ProbabilityHow likely it is that a particular outcome will be the result of a random circumstance

The Relative Frequency Interpretation of Probability

In situations that we can imagine repeating many times, we define the probability of a specific outcome as the proportion of times it would occur over the long run -- called the relative frequencyof that particular outcome.

Example: Probability of Male versus Female Births

Long-run relative frequency of males born in a State is about 0.512 (512 boys born per 1000 births)

Table provides results of simulation: the proportion is far from .512 over the first few weeks but in the long run settles down around .512.

Determining the Relative Frequency(Probability) of an Outcome

Method 1:Make an Assumption about the Physical World (there is no bias)

A Simple LotteryChoose a three-digit number between 000 and 999. Player wins if his or her three-digit number is chosen. Suppose the 1000 possible 3-digit numbers (000, 001, 002, 999) are equally likely.In long run, a player shouldwin about 1 out of 1000 times. Probability = 0.001 of winning.This does not mean a player will win exactly once in every thousand plays.

Determining the Relative Frequency(Probability) of an Outcome

Method 2: Observe the Relative Frequency of random circumstances

The Probability of Lost Luggage“1 in 176 passengers on XYZ airline carriers will temporarily lose their luggage.”This number is based on data collected over the long run. So the probability that a randomly selected passenger on a XYZ carrier will temporarily lose luggage is 1/176 or about 0.006.

Ways to express the relative frequency of lost luggage:

• The proportion of passengers who lose their luggage is 1/176 or about 0.006 (6 out of 1000).

• About 0.6% of passengers lose their luggage.

• The probability that a randomly selected passenger will lose his/her luggage is about 0.006.

• The probability that you will lose your luggage is about 0.006.

Last statement is notexactlycorrect – your probability depends on other factors (how late you arrive at the airport, etc.).

Personal probability of an event = the degree to which a given individual believes the event will happen. Sometimes subjective probability used because the degree of belief may be different for each individual.

• Restrictions on personal probabilities:

• Must fall between 0 and 1 (or between 0 and 100%).

• Must be coherent (logical and well organized).

Sample space: collection of unique, nonoverlapping possible outcomes of a random circumstance.

Simple event: one outcome in the sample space; a possible outcome of a random circumstance.

Event: a collection of one or more simple events in the sample space; often written as A, B, C, and so on.

P(A) = probability of the event A

• Conditions for Valid Probabilities

• Each probability is between 0 and 1.

• The sum of the probabilities over all possible simple events is 1.

Equally Likely Simple EventsIf there are k simple events in the sample space and they are all equally likely, then the probability of the occurrence of each one is 1/k.

Random Circumstance: A three-digit winning lottery number is selected.Sample Space: {000,001,002,003, . . . ,997,998,999}. There are 1000 simple events.Probabilities for Simple Event: Probability any specific three-digit number is a winner is 1/1000. Assume all three-digit numbers are equally likely.

Event A = last digit is a 9 = {009,019, . . . ,999}. Since one out of ten numbers in set, P(A) = 1/10.

Event B = three digits are all the same = {000, 111, 222, 333, 444, 555, 666, 777, 888, 999}. Since event B contains 10 events, P(B) = 10/1000 = 1/100.

One event is the complement of another event if the two events do not contain any of the same simple events and together they cover the entire sample space.

Notation: AC represents the complement of A.

Note: P(A) + P(AC) = 1

Example:A Simple Lottery (cont)A = player buying single ticket wins AC = player does not winP(A) = 1/1000 so P(AC) = 999/1000

Two events are mutually exclusive, or equivalently disjoint,if they do not contain any of the same simple events (outcomes).

Example; A Simple LotteryA = all three digits are the same. B = the first and last digits are differentThe events A and B are mutually exclusive (disjoint), but they are not complementary.

• Two events are independent of each other if knowing that one will occur (or has occurred) does not change the probability that the other occurs.

• Two events are dependent if knowing that one will occur (or has occurred) changes the probability that the other occurs.

The definitions can apply either … to events within the same random circumstance or to events from two separate random circumstances.

Example Independent Events : Winning a Free Lunch

• Customers put business card in restaurant glass bowl.

• ABC and XYZ put a card in two consecutive wks.

Event A = ABC win in week 1.

Event B = XYZ wins in week 1.

Event C = XYZ wins in week 2.

• Events A and B refer to the same random circumstance and are not independent.

• Events A and C refer to different random circumstances and are independent.

Example: Dependent Events

Event A = ABC is selected to answer Question 1.

Event B = ABC is selected to answer Question 2.

Events A and B refer to different random circumstances,

but are A and B independent events?

• P(A) = 1/50. (Assuming that there are 50 cards in the box)

• If event A occurs, her name is no longer in the bag; P(B) = 0.

• If event A does not occur, there are 49 names in the bag (including Alicia’s name), so P(B) = 1/49.

Knowing whether A occurred changes P(B). Thus, the events A and B are not independent.

Conditional probability of the event B, given that the event A occurs, is the long-run relative frequency with which event B occurs when circumstances are such that A also occurs; written as P(B|A).

P(B) = unconditional probability event B occurs.

P(B|A) = “probability of B given A”

= conditional probability event B occurs given that we know A has occurred or will occur.

Probability an Event Does Not Occur

Rule 1 (for “not the event”): P(AC) = 1 – P(A)

Example Probability a Stranger Does NotShare BirthdaysP(next stranger you meet shares your birthday) = 1/365 (low P).P(next stranger you meet will not share your birthday) = 1 – 1/365 = 364/365 = 0.9973 (high P)

Rule 2 (addition rule for “either/or”):

Rule 2a (general): P(A or B) = P(A) + P(B) – P(A and B)

Rule 2b (for mutually exclusive events): If A and B are mutually exclusive events, P(A or B) = P(A) + P(B)

Example:Probability That Either of Two Events Happen

Ram is off to college. There are 1000 male students. Ram hopes his roommate will not like to party and not snore.

A = likes to party P(A) = 250/1000 = 0.25

B = snores P(B) = 350/1000 = 0.35

A +B = snores and party P(A and B) = 150/1000 = 0.15

Probability that Ram will be assigned a roommate who either

likes to party or snores, or both is: P(A or B)

= P(A) + P(B) – P(A and B) = 0.25 + 0.35 – 0.15 = 0.45

So the probability his roommate is acceptable is 1 – 0.45 = 0.55

Probability That Two or More Events Occur Together

Rule 3 (multiplication rule for “and”):

Rule 3a (general): P(A and B) = P(A)P(B|A)

Rule 3b (for independent events): If A and B are independent events, P(A and B) = P(A)P(B)

Extension of Rule 3b (for > 2 indep events): For several independent events,P(A1 and A2 and … and An) = P(A1)P(A2)…P(An)

P of B given A

Example:Probability That Two or More Events Occur Together

For 9th graders, 22.9% of the boys and 4.5% of the girls admitted they gambled at least once a week during the previous year. The population consisted of 50.9% girls and 49.1% boys. What is P that a randomly selected student will be a male who also gambles? (event A = male selected; even B = a weekly gambler is selected; events A and B are dependent [Rule 3a (general multiplication - P(A)P(B|A)]

Event A = male Event B = weekly gambler

P(A) = 0.491 P(B|A) = 0.229

P(maleandgambler) = P(A and B) = P(A)P(B|A) = (0.491)(0.229) = 0.1124

Rule 4 (conditional probability):

P(B|A) = P(A and B)/P(A)

And it is also true that:

P(A|B) = P(A and B)/P(B)

Example:Determining a Conditional Probability

If we know ABC is picked to answer one of three questions, what is the probability it was the first question?

A = ABC selected to answer Question 1, P(A) = 1/50

B = ABC is selected to answer any one of the questions, P(B) = 3/50

Since A is a subset of B, P(A and B) = 1/50

P(A|B) = P(A and B)/P(B) = (1/50)/(3/50) = 1/3

Students sometimes confuse the definitions of independent and mutually exclusive events.

• When two events are mutually exclusive and one happens, it turns the probability of the other one to 0.

• When two events are independent and one happens, it leaves the probability of the other one alone.

Hints and Advice for Finding Probabilities

• P(A and B): define event in physical terms and see if know probability. Else try multiplication rule (Rule 3).

• Series of independent events all happen: multiply all individual probabilities (Extension of Rule 3b)

• One of a collection of mutually exclusive events happens: add all individual probabilities (Rule 2b extended).

• Check if probability of complementeasier, then subtract it from 1 (applying Rule 1).

• None of a collection of mutually exclusive events happens: find probability one happens, then subtract that from 1.

• Conditional probability: define event in physical terms and see if know probability. Else try Rule 4 or next bullet as well.

• Know P(B|A) but want P(A|B): Use Rule 3a to find P(B) = P(A and B) + P(AC and B), then use Rule 4.

Step 1: List each separate random circumstance involved in the problem.

Step 2: List the possible outcomes for each random circumstance.

Step 3: Assign whatever probabilities you can with the knowledge you have.

Step 4: Specify the event for which you want to determine the probability.

Step 5: Determine which of the probabilities from step 3 and which probability rules can be combined to find the probability of interest.

• Let S be the sample space and E1, E2,…, En are n mutually exclusive events then

• In other words “ An event is known to have proceeded from one of the mutually exclusive causes whose probabilities are P1, P2,…,P3”. Further p1,p2,…,pn be the respective probabilities

that when one of the n causes exists, the event will then have followed. The probability that the event posses the mth cause is given by

Example: have followed. The probability that the event posses the mth cause is given by

A doctor is to visit a patient. From the past experience, it is known that the probabilities that he will come by train, bus, scooter or by other means of transport are respectively 3/10, 1/5, 1/10 and 2/5. The probabilities that he will be late are ¼, 1/3 and 1/12 if he comes by trains , bus, scooter respectively, but if he comes by other means of transport, then he will not be late. When he arrives he is late. What is the probability that he comes by train?

Solution: have followed. The probability that the event posses the mth cause is given by

Let P1 = Prob. that doctor comes by train = 3/10

P2= Prob. that doctor comes by bus = 1/5

P3= Prob. that doctor comes by scooter = 1/10

P4= Prob. that doctor comes by other means = 2/5

Also, p1 = Prob. that doctor will be late if he comes by train = ¼

p2= Prob that the doctor will be late if he comes by bus =1/3

Solution: have followed. The probability that the event posses the mth cause is given by

p3= Prob that the doctor will be late if he comes by scooter =1/12

p4= Prob that doctor will be late if he comes by other means = 0

By Bay’s theorem

The Prob that doctor arriving late comes by train is