Presentation on theme: "Chapter 14 From Randomness to Probability. Random Phenomena ● A situation where we know all the possible outcomes, but we don’t know which one will or."— Presentation transcript:
Random Phenomena ● A situation where we know all the possible outcomes, but we don’t know which one will or has happened.
Probability Probability- an event’s long run relative frequency Outcome- the result of the trial Event- a combination of outcomes Independent (trial)- the outcome of the trial doesn’t change or influence the outcome of another
The Law of Large Numbers ● The long run relative frequency of repeated independent events gets closer and closer to the true relative frequency as the number of trials increase. ● The law of large numbers does not compensate for whatever happened in the past.
Personal Probability ● Personal probabilities are ones we express that are not based on long run relative frequencies. ● They do not display the consistency we need for probabilities, so stick with formally defined probabilities.
Formal Probability ● There are two requirements for a formal probability -a probability is a number between 0 and 1 -For any event A, 0 ≤ P(A) ≤ 1 ● The “Something has to Happen Rule” -the probability of the set of all possible outcomes must equal 1
Formal Probability Cont. ● Complement Rule -The set of outcomes that are not in the event A is called a complement of A, denoted A C -The probability of an event occurring is 1 minus the probability that it doesn’t occur: P(A) = 1 – P(A C )
Formal Probability Cont. ● Addition Rule -Events that have no outcomes in common are called disjoint. -P(A or B) = P(A) + P(B), provided that A and B are disjoint
Formal Probability Cont. ● Multiplication Rule -Used for two independent events -P(A and B) = P(A) x P(B), provided that A and B are independent and have probabilities greater than 0.
Problem #21 As mentioned in the chapter, opinion polling organizations contact their respondents by sampling random telephone numbers. Although interviewers can reach about 76% of US households, the percentage of those contacted who agree to cooperate with the survey has fallen from 58% in 1997 to 38% in 2003 Each household, of course, is independent of others.
Solving 21 A.What is the probability that the next household on the list will be contacted, but will refuse to cooperate? -Well, we know the probability that they do cooperate. Which is 58% in 97, and 38% in 2003. We must also incorporate only 76% of household. -Subtract the chance of cooperation in 2003 from 1. 1-0.38=0.62 -And then multiply that by the amount of households contacted. The result will be the probability. 0.62*0.76=0.4712
Solving 21 B.What is the probability (in 2003) of failing to contact a household or of contacting the household but not getting them to agree to interview? -The question asks OR. So We just need to take the probability of both events occuring, multiply, and use the sum to subtract from 1. 1-(0.76)(0.38)= 0.7112
Solving 21 C.Show another way to solve part B (1-0.76)+0.76(1-0.38)=0.7112 -Take the probability of the house being contacted and subtract it from 1 to find houses that were not contacted. (1- 0.76) -Add the result to the probability the house being contacted, but refused to cooperate 0.24+0.76(1-0.38) -You see 0.76 being multiplied on the parentheses in order to only take houses contacted into account.
Problem #23 The Masterfoods company says that before the introduction of purple, yellow candies made up 20% of their plain M&Ms, red another 20%, and orange, blue, and green each made up 10%. The rest were brown.
Solving 23 A.If you pick an M&M at random, what is the probability that... 1.it is brown?.3, divide the 30% by 100% of the M&Ms 1.it is yellow or orange?.3, yellow-20% orange-10% 20+10=30% 30/100=.3 1.it is not green?.9, 10% are green, 100-10=90% (not green) 90/100=.9 1.it is striped? None of the M&Ms were striped therfore the probability of a striped one being picked will be 0
Solving 23 B.If you pick three M&M’s in a row what is the probability that… 1.they are all brown?.027, the probability that you will pick a brown one is.3, therefore to find the probability that you will pick three in a row you use this equation (.3)(.3)(.3)=.027 1.the third one is the first one that’s red?.128, probability not red=.8 red=.2 (.8)(.8)(.2)=.128 1.none are yellow?.512, probability not yellow=.8 (.8)(.8)(.8)=.512 1.at least one is green?.271, Subtract the probability of none of the M&M’s being green by one (.9)(.9)(.9)=.729 1-.729=.271