Chapter 14 From Randomness to Probability Random Phenomena

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 has happened.

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 AC -The probability of an event occurring is 1 minus the probability that it doesn’t occur: P(A) = 1 – P(AC)

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)=. 027 1. the third one is the first one that’s red? . 128, probability not red=. 8 red=. 2 (. 8)(. 2)=. 128 1. none are yellow? . 512, probability not yellow=. 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)=. 729 1 -. 729=. 271