Chapter 4
CHAPTER 4
Section 4.1
Event- A set of outcomes of a probability experiment.
Simple Event- An event that can not be broken down into simpler terms.
Sample Space- Set of all possible outcomes of a probability experiment.
Classical Approach to Probability (used for equally likely outcomes)
P(A)= number of ways A can occur/number of simple events
Complement of an Event A --> A' (book uses an A with a bar on top)
A' =every outcome not in A
All probabilities will be between 0 and 1, inclusive. 0 < P(x) < 1
4-1 # 9/10 Assume that 50 births are randomly selected. Use subjective judgment to describe the given number of girls as (a) significantly low, (b) significantly high, (c) neither significantly low nor high.
#9) 47 girls
significantly high, since we wouldn't expect near that many girls in 50 births.
#10) 26 girls
neither, since we would expect about the same number of boys as girls.
Table 4-1 Results from Drug Tests of Job Applicants
Positive test result | Negative test result
Subject uses drugs 45(true pos) 5 (false neg) row total=50
Subject does not use drugs 25(false pos) 480 (true neg) row total=505
column total=70 column total =485 grand total=555
4-1 #22
Find the probability of selecting someone who got a result that is a false positive.
P(false positive)=25/555 (25 false positives in the total of 555)
Who would suffer from a false positive result? Why? The non drug users, they tested positive for drugs when they are not using. It could cost them a chance at the job.
4-1 #24
Find the probability of selecting someone who does not use drugs. Does the result appear to be reasonable as an estimate of the proportion of the adult population that does not use drugs?
P(does not use drugs)=505/555 second row total is the number of non users over the grand total.
4-1 #28
Guessing Birthdays. On their first date, Kelly asks Mike to guess the date of her birth, not including the year.
a) What is the probability that Mike will guess correctly?
P(guessing bday)=1/365 only one day a year is her bday.
b) Would it be unlikely for him to guess correctly on his first try?
Yes, not much of a chance.
c) If you were Kelly, and Mike did guess correctly on his first try, would you believe his claim that he made a lucky guess, or would you be convinced that he already knew when you were born?
He already knew! Maybe a google search before the date?
d) If Kelly asked Mike to guess her age, and Mike’s guess is too high by 15 years, what is the probability that Mike and Kelly will have a second date?
Just a silly one...very low!
Odds:
A modified roulette wheel has 40 slots. One slot is 0, another is 00, and the others are numbered 1 through 38, respectively. You are placing a bet that the outcome is an odd number. (In roulette, 0 and 00 are neither odd nor even.)
a. What is your probability of winning?
The probability of winning is 19/40 (40 slots, 19 being an odd number)
b. What are the actual odds against winning?
The actual odds against winning are 21:19 (21 ways to not win and 19 ways to win)
c. When you bet that the outcome is an odd number, the payoff odds are 1:1. How much profit do you make if you bet $16 and win?
If you win, the payoff is $16 (1:1 means $1 won for each dollar bet)
d. How much profit should you make on the $16 bet if you could somehow convince the casino to change its payoff odds so that they are the same as the actual odds against winning?
21/19 * $16 = $17.68
SECTION 4-2 Addition and Multiplication Rules
Addition Rule: KEY WORD FOR USING THE ADDITION RULE IS "OR"!!
P(A or B)= P(A) + P(B) – P (A and B)
P(A or B)= P(A) + P(B) if A and B are mutually exclusive (disjoint)
Mutually Exclusive / Disjoint – Two events that can not occur at the same time.
Example of Mutually Exclusive / Disjoint events:
Event A-riding a bike
Event B-driving a car
These events would be Mutually Exclusive / Disjoint as you can not do them at the same time.
Event A-driving a car
Event B-texting
These event would NOT be Mutually Exclusive / Disjoint as you can drive a text at the same time (even though it is illegal and not advised)
Table 4-1 Results from Drug Tests of Job Applicants
Positive test result | Negative test result
Subject uses drugs 45(true pos) 5 (false neg) row total=50
Subject does not use drugs 25(false pos) 480 (true neg) row total=505
column total=70 column total =485 grand total=555
Example A)
If one of the test subjects is randomly selected, find the probability that the subject had a positive test result or negative test result.
Note: These are mutually exclusive (disjoint) events because you can not have a positive and negative test result at the same time...use the second formula.
P(positive test OR negative test)=P(positive test)+P(negative test)=70/555 +485/555= 555/555=1
Example B)
If one of the test subjects is randomly selected, find the probability that the subject had a negative test result or does not use drugs.
These are not Mutually Exclusive / Disjoint since you can have a negative test and not use drugs.
P(A or B)= P(A) + P(B) – P (A and B)
P(negative test OR does not use drugs)=P(negative test) + P(does not use drugs) - P(negative test and does not use)
=485/555 + 505/555 - 480/555= 510/555
Note: We are counting the 480 twice...once with the negative test and once with the does not use drugs. Therefore, we need to subtract them out once.
Video of Addition Rule Example
Multiplication Rule:KEY WORD FOR USING THE MULTIPLICATION RULE IS "AND"!!
P(A and B)=P(A) * P(B|A)
P(A and B)=P(A) * P(B) if A and B are independent.
Independent Events- Two events such that one occurring does not affect the occurrence of the other.
P(B|A) – probability of B given A has already occurred.
Example of Independent events:
Event A-Having a baby girl
Event B-rolling a 4 on a die
These two events are independent since the occurrence of one has no effect on the other.
Example of dependent events:
Assume you have a bag of candy with 4 Red candies and 4 Blue candies
Event A-drawing a red candy
Event B-drawing a second red candy
The probability of the first candy being red (not looking in the bag, just randomly selecting) is 4/8. If the first candy is red, then the probability the second is red would be 3/7 (one less candy and one less red...since we ate the first red candy as soon as we took it out of the bag!).
Event A-drawing a blue candy
Event B-drawing a second red candy
The probability of the first candy being blue (not looking in the bag, just randomly selecting) is 4/8. If the first candy is blue, then the probability the second is red would be 4/7 (one less candy...since we ate the first blue candy as soon as we took it out of the bag!).
Notice that the probability has changed for the second candy being red depending on what was taken out on the first grab....this is dependent!
Table 4-1 Results from Drug Tests of Job Applicants
Positive test result | Negative test result
Subject uses drugs 45(true pos) 5 (false neg) row total=50
Subject does not use drugs 25(false pos) 480 (true neg) row total=505
column total=70 column total =485 grand total=555
Example C)
If 2 of the 555 test subjects are randomly selected, find the probability that they both had false positive results.
i) Assume that the 2 selections are made with replacement
WITH REPLACEMENT...that means INDEPENDENT!!
P(A and B)=P(A) * P(B)
=P(1st is a false positive AND the 2nd is a false positive)
=P(1st false positive)*P(2nd false positive)
=25/555*25/555
=.002029
ii) Assume that the 2 selections are made without replacement.
WITHOUT REPLACEMENT...that means DEPENDENT!! This is like the candy example above.
P(A and B)=P(A) * P(B|A)
=P(1st is a false positive AND the 2nd is a false positive)
=P(1st is a false positive)*P(2nd false positive| 1st false positive)
note:P(2nd false positive| 1st false positive) just means assume the first one taken out was a false positive.
=25/555 * 24/554 (one less false positive and one less in the total)
=.00195
Video of Multiplication Rule Example
Acceptance Sampling. With one method of a procedure called acceptance sampling, a sample of items is randomly selected without replacement and the entire batch is accepted if every item in the sample is found to be okay.
4-2 #28
NOAA inspects seafood that is to be consumed. The inspection process involves selecting seafood samples from a larger “lot”. Assume a lot contains 2875 seafood containers and 288 of these containers include seafood that does not meet inspection requirements. What is the probability that 3 selected container samples all meet requirements and the entire lot is accepted based on this sample? Does this probability seem adequate?
total=2875
bad=288
good=2587
This would be a dependent case as sample are done without replacement!
P(all 3 selected are good)=(2587/2875)*(2586/2874)*(2585/2873)=.7285
note: at each step we had one less good container and one less in the total..."count down" at each step.
4-2 #29 (Find the probabilities and indicate when the “5% guideline for cumbersome calculations” is used.). In a study of helicopter usage and patient survival, results were obtained from 47,637 patients transported by helicopter and 111,874 patients transported by ground.
a) If 1 of the 159,511 patients in the study is randomly selected, what is the probability that the subject was transported by helicopter?
P(transported by helicopter)= 47,637/159,511
b) If 5 subjects in the study are randomly selected without replacement, what is the probability that all of them were transported by helicopter?
Ok, since this is without replacement this is technically dependent and we should use the "count down" technique as we did in #28 above. The issue with that is that we are doing that 5 times and it gets annoying...but there is a trick to avoid that annoying calculation! The “5% guideline for cumbersome calculations”. This states that if less that 5% of the total is selected, we can treat this as independent. Ie. we are selecting 5 out of 159511 which is less that 5% so we can treat as independent.
P(all 5 being transported by helicopter)=(47,637/159,511)^5 =.00238
Section 4-3 Complements, Conditional Probability, and Bayes’ Theorem
Complements:
A' = Every element not in A
Complement rules:
P(A) = 1- P(A') note the book will use A with a line above it rather than A '
P(A) = 1- P(A')
P(A) + P(A') = 1
KEY PHASE FOR COMPLEMENT RULE..."AT LEAST ONE or AT MOST x"
Example: If event A is rolling a 4 on a die, then A' would be everything other than a 4...rolling a 1, 2, 3, 5, or 6.
4-3 #5 Find the probability that when a couple has three children, at least one of them is a girl. (Assume boys and girls are equally likely.)
Complement rule since is states "at least one"
P(at least one girl)=1-P(no girls)=1-P(all 3 are boys)=1-(1/2)^3 =7/8
Video of finding "at least one" and the benefit of the complement rule
Conditional Probability:
P(B|A) = P(a and B)/P(A) P(B|A)…”B given that A has already occurred”
Key phase for conditional probability... "given that".
In an experiment to study the effects of using four quarters or a $1 bill, college students were given either four quarters or a $1 bill and they could either keep the money or spend it on gum.
Purchased Gum Kept the Money totals
Students Given Four Quarters 27 16 43
Students Given a $1 bill 12 34 46
totals 39 50 89
4-3 #13 Denomination Effect
a) Find the probability of randomly selecting a student who spent the money, given that the student was given four quarters.
P(B|A) =P(A and B) /P(A)
= P(spent $ | 4 quarters) = P(4 quarters and spent $)/ P(4 quarters) = (27/89) / (43/89) =27/43
I've always had trouble following the above formula and I recommend looking at this in a different way. We know that the student was given 4 quarters so it has to come from the first row of data. Out of the total of 43 in that group, how many spent the money? 27 so, 27/43 as we found above! We can use this short cut anytime we see the "given that" phrase.
b) Find the probability of randomly selecting a student who kept the money, given that the student was given four quarters.
Using our short cut, we know that it comes from the 43 that were given the 4 quarters...out of those 43, 16 kept the money. So, the probability is 16/43
Video of conditional probability (short cut)
4-3 #24 Composite Water Samples. The Fairfield County Department of Public Health tests water for presence of E. coli bacteria. To reduce laboratory costs, water samples from 10 public swimming areas are combined for one test, and further testing is done only if the combined sample tests positive. Based on past results, there is a 0.005 probability of finding E. coli bacteria in a public swimming area. Find the probability that a combined sample from 10 public swimming areas will reveal the presence of E. coli bacteria. Is that probability low enough so that further testing of the individual samples is rarely necessary?
P(E. coli)=0.005
P(NO E. coli)=.995
P(sample tests positive)=P(at least one of the 10 samples have E. coli)
=1-P(none have E. coli)...from the complement rule.
=1-(.995)^10...probability of NO E. coli is .955 and we want all 10 not to have it
=.0489
Section 4-4 Counting
Fundamental Counting Rule (FCR): If 1st event can occur M ways and 2nd event can occur N ways, together they can occur M*N ways.
Example: If we have a die that has 6 outcomes and a coin that has 2 outcomes, there would be 2*6=12 outcomes together:
1T 1H
2T 2H
3T 3H
4T 4H
5T 5H
6T 6H
4-4# 5 ATM Pin Numbers. A thief steals an ATM card and must randomly guess the correct pin code that consists of four digits (each 0 through 9) that must be entered in the correct order. Repetition of digits is allowed. What is the probability of a correct guess on the first try?
Each position has 10 choices. 4 digits total.
10*10*10*10=10,000 possibilities.
P(guessing on first try)=1/10,000
Video Example of the Fundamental Counting Rule
Factorial Rule: n different items can be arranged n! ways
Example: family picture with 5 family members
5! =5*4*3*2*1=120 different possible arrangements
Use the ! key on your calculator, do not multiply these out!
Video using a calculator to find factorials (NOTE: This is using a graphing calculator but just about any calculator has these functions, try googling your model number and ! to find directions for your calculator)
4-4 #9 Grading Exams. Your professor has just collected eight different statistics exams. If these exams are graded in random order, what is the probability that they are graded in alphabetical order of the students that took the exam?
number of possibilities 8!=40320
P(alphabetical order)=1/40,320 not likely!!
Permutation Rule (Order matters)
nPr =n!/ (n-r)! (When all items are different)
n!/ (n1! * n2! *…nk!) (When some items are identical to others)
Video of Factorial Rule and Permutation Rule (with some items identical) This video is both and example like number 9 above and 22 below.
4-4 #11 Scheduling Routes. A presidential candidate plans to begin her campaign by visiting the capitals of 5 of the 50 states. If the five capitals are randomly selected without replacement, what is the probability that the route is Sacramento, Albany, Juneau, Hartford, and Bismarck, in that order?
order matters since changing the cities around would completely change the trip...permutation.
50P5=254,251,200 options for selecting 5 out of 50 capitals. Use the nPr key on your calculator, do not figure this out with the formula above.
P(Sacramento, Albany, Juneau, Hartford, and Bismarck, in that order)=1/254,251,200
Video with several examples of combinations and permutation (like number 11 above and 29 below) (NOTE: This is using a graphing calculator but just about any calculator has these functions, try googling your model number and nCr or nPr to find directions for your calculator)
4-4 # 22 A classic counting problem is to determine the number of different ways that the letters of “Mississippi” can be arranged. Find that number.
n!/ (n1! * n2! *…nk!) (When some items are identical to others)
11!/(4! * 4! * 2!) = 34,650 arrangements. We need to use the ! key on your calculator multiple times for this problem.
Combination Rule (Order does NOT matter)
nCr = n!/((n-r)! r!)
4-4 #29 Mega Millions. The Mega Millions is run in 44 states. Winning the jackpot requires that you select the correct five different numbers between 1 and 75 and, in a separate drawing, you must also select the correct single number between 1 and 15. Find the probability of winning the jackpot. How does the result compare to the probability of being struck by lightning in a year, which the National Weather Service estimates to be 1/960, 000?
order does not matter as you either have the winning numbers or not, order is not important
75C5 * 15C1 (pick 5 out of 75 and 1 out of 15). Use the nCr key on your calculator, not the formula above.
= 17,259,390*15
=258,890,850
P(winning)=1/258,890,850
So, it's approximately 270 times more likely to be struck by lightning than winning the lotto!!
Video of Combinations and Permutations (and how to distinguish the difference)