Stat 302 - Fall 2020Week In Review

Week ]4: More Probability

Problem 1. Probability is a field of mathematics that allows us to think about random eventsand quantify how often particular outcomes will occur. There are two main ways we can thinkabout probability. The first is known as the frequentist approach. This approach looks atprobability as a long term proportion: if we repeated this trial/experiment an infinite number oftimes, how often would a particular event occur. Let’s begin by thinking of an unfair coin. Thiscoin is weighted such that heads will show up 3 times more often than tails.

a. If heads will show up 3 times more often than tails, what proportion of the time do you expectto get heads? What proportion of the time do you expect to get tails.

b. What will the proportion of tails be after one flip? What are the possible values?c. What happens to the proportion of tails as we flip the coin more and more times?

Problem 2. Estimating probabilities by looking at long term proportions is a wonderful idea inmany cases, but it is not always feasible. Sometimes, we have to use what is called a subjective orpersonal probability. For instance, if you wanted to know what is the probability of you gettingan A in a particular class, you wouldn’t be able to look at the long term proportion to estimatethis. Can you think of another example where you would need to use a subjective probability?

Shelby Cummings Week ]4 Page 1 of 9

a. PCH)-- O -75

c. get closer to whatPIT) -- 0.25 we expect to happen

b. Is either going to 0.25to be 0 or 1 .

--

inventing- airplanes

Stat 302 - Fall 2020Week In Review

Problem 3. While the most common die has six sides, there are a number of other varieties ofdie that are used in various games. One such die is the four-sided die. Assume we have two fairfour-sided die, each labeled with the numbers from one to four.

a. What do we mean by a fair die?b. What is the sample space for rolling one die?c. What is the sample space for rolling two die?d. Assume you roll both die and add them together. What is the sample space?e. Assume you roll both die and add them together. What is the probability the sum is greater

than 6? It may be a good idea to use the sample space from part c to solve this.

Shelby Cummings Week ]4 Page 2 of 9

DE E

-

- g

a.PTOB of each outcome is equalPll) =p (2) =p(3) =p(4)

= 0.25

b. S = { 1,2 ,3,4}

C.

S = { ( l , D ,(1,2) , 11,3), 11,4

),

12,1) , 12,2) , 12,3),(2,4)

(3,1) , (3,2),13,3) ,(3,4) .

'

(4,1) , 14,2) ,14,3), 14,4)}

d. S = { 2.3,4 , 5,617,83

e . P (sums6) =p(sum --7 U sum = 8)

=p(sum =7) t PCsum = 8)= ÷ t IT= ÷ f-0.187€

Stat 302 - Fall 2020Week In Review

Problem 4. Many times, we organize our information about a random event by creating a tableor diagram that describes the probability distribution. The table below shows the probabilitydistribution of Educational Attainment for Americans 25 years and older in 2019, as reported bythe Census Bureau.

x None No HS Diploma HS Diploma Some College Assoc. Bach Bach+P(x) 0.003 0.096 0.281 0.157 0.103 0.225 0.135

a. What type of variable is educational attainment?b. What two properties must be true for a probability distribution to be valid? Is this a valid

probability distribution?c. What is the probability that a randomly selected person does not have a high school diploma?d. What is the probability that a randomly selected person has some college degree?

Shelby Cummings Week ]4 Page 3 of 9

some education (no degree) (mdoa§g¥ue)

a . categorical -ordinalb. ① sum = 1

② each prob .is between O da I✓

yes , this is a valid prob. dist

'

n

C . P(didn't graduate HS) = P(none U no

Its dip)

=P (none) t P(some ed, noHS diploma)

= 0.003 t O .096=0.0990d. PC some college degree

) =

Plasse U bachU bacht)

=P (assoc)t PLbach)t P(bacht)

= 0.103 t 0.225 to . 135=10.4637

Stat 302 - Fall 2020Week In Review

Problem 5. Let’s continue looking at the 2019 Educational Attainment data, however, now let’slook at it split up by gender.

None No HS Diploma HS Diploma Some College Assoc. Bach Bach+ TotalMales 0.002 0.048 0.141 0.075 0.045 0.107 0.063 0.482Females 0.002 0.047 0.140 0.082 0.058 0.118 0.072 0.518Total 0.004 0.095 0.281 0.157 0.103 0.225 0.135 1

a. Are having an associates degree and being female mutually exclusive?b. What is the probability that a randomly selected person has an associates degree or is a

female?c. What is the probability that a randomly selected person has an associates degree and is a

female?d. What is the probability that a randomly selected female has an associates degree?e. What is the probability that a randomly selected person has an associates degree if they are

a male?f. Does it appear as though gender and educational attainment are dependent or independent?Why?

Shelby Cummings Week ]4 Page 4 of 9

plAUB) -- PLA) t PCB) - planB)

someeduc. no degree

a. mutually exclusive =disjointassociates de female - not disjoint ,

because

both can happen at the sametime

b . P(assoc U female) =p(assoc)t poemale)- p(assoc n female)

=0.103+0.518-0.058=0.55C. Plassoc A female) -40.058J

Stat 302 - Fall 2020Week In Review

Problem 5. Let’s continue looking at the 2019 Educational Attainment data, however, now let’slook at it split up by gender.

None No HS Diploma HS Diploma Some College Assoc. Bach Bach+ TotalMales 0.002 0.048 0.141 0.075 0.045 0.107 0.063 0.482Females 0.002 0.047 0.140 0.082 0.058 0.118 0.072 0.518Total 0.004 0.095 0.281 0.157 0.103 0.225 0.135 1

a. Are having an associates degree and being female mutually exclusive?b. What is the probability that a randomly selected person has an associates degree or is a

female?c. What is the probability that a randomly selected person has an associates degree and is a

female?d. What is the probability that a randomly selected female has an associates degree?e. What is the probability that a randomly selected person has an associates degree if they are

a male?f. Does it appear as though gender and educational attainment are dependent or independent?Why?

Shelby Cummings Week ]4 Page 4 of 9

NAIB)=P¥BPLB)

=Plassochfemale)

d. Plassoctfemale)-=0.058

Ptfemate)

⇒= 0.112mL

E. Plassocl male)=pLassoon

male)

= Lig =1⇒÷)f.We aren't sure , PL assoc)

is dif .

depending on gender, it is a small dlf.

Stat 302 - Fall 2020Week In Review

Problem 6. Many times when we are discussing probabilities about diseases, we talk about therisk and the odds. The risk is the same as the probability, while the odds is the probabilitydivided by one minus the probability. One of the most common genetic disorders in the UnitedStates is Down Syndrome. If an individual has Down Syndrome, there is an increased chancethat they will have a heart defect. Approximately 47% if infants born with Down Syndrome alsohave a heart defect?

a. If a child has Down Syndrome, what is the risk of them having a heart defect?b. If a child has Down Syndrome, what is the odds of them having a heart defect? What does

this mean?

Problem 7. Many times when we are looking at probability problems, we are not asking aboutjust one event, but rather a combination of events. Let’s look at an example where we are lookingat two di↵erent events: A student is taking a history class that has one midterm and one finalexam. They believes that the chance of them getting an A on the midterm is 50%. They alsobelieve the chance of them getting an A on the final exam is 50%. The student states that theirchance of getting an A on at least one of the two tests is 100%. Is their reasoning correct? Whyor why not? If it is incorrect, how would you solve for the probability of them getting an A onat least one of the two tests?

Shelby Cummings Week ]4 Page 5 of 9

Risk =pOdds -- Pl- p

•

°

a. P(Heart defect 1 Down syndrome)= 0.47

b. ODDS = = fj.tt#--Oo?sI3--o.887&Children with Down syndrome are more

likely to not have a heartdefect

* A on MT 9 A on

Plget an A on MT) -- 0.50 the final are notdisjoint

Plget an A onfinal) -- 0.50

PLA on MT VA onfinal) = PLAonMT)t PCA onfinal)

-PCA on both)

NO , they forgotto subtract the intersection

Stat 302 - Fall 2020Week In Review

Problem 8. Let’s look at another example of a probability question involving two events. Asoftware company provides an email filtering service to protect email users from spam. Thecompany has advertised that their software is 95% accurate. This could mean one of four things.Listed below are the four things this could mean. Write out each as a conditional probabilitystatement.

a. 95% of the blocked emails are spamb. 95% of spam emails are blockedc. 95% of the valid emails are allowed throughd. 95% of the emails allowed through are valid

Shelby Cummings Week ]4 Page 6 of 9

S -- spam

Sc -- valid

A.PH/B)-- 0.95B-- Blocked

BC = allowed

b. PIB 151=0.95

c. PCB' 154=0.95

d. P(54139=0.95

Stat 302 - Fall 2020Week In Review

Problem 9. Now let’s continue looking at conditional probability. Based on past drug testingof air tra�c controllers, the FAA reports that the probability of drug use at any given time isapproximately 0.007. The FAA uses a particular test to determine if the air tra�c controllersare currently using drugs is 96% sensitive and 93% specific.

a. What is the probability of a positive test?b. If a test is positive, what is the probability that the individual is actually using drugs?

Shelby Cummings Week ]4 Page 7 of 9

prev -_ PLDT -_ 0.007 PIT'T ?

Sens-_PCTtlD'7=0.96 pLDtITtJ= ?

spec-- PLT - ID-1=09710.007310-96)

-_0.00672pt 1-0.96

¥ *t

Tt (0.993)(0.07)

+0.007 = 0.06951

Eres- T

a. PITT)=p(TtnDt) t PLTTND

-)

-_0.00672+0.06951=10.076237

Stat 302 - Fall 2020Week In Review

Problem 9. Now let’s continue looking at conditional probability. Based on past drug testingof air tra�c controllers, the FAA reports that the probability of drug use at any given time isapproximately 0.007. The FAA uses a particular test to determine if the air tra�c controllersare currently using drugs is 96% sensitive and 93% specific.

a. What is the probability of a positive test?b. If a test is positive, what is the probability that the individual is actually using drugs?

Shelby Cummings Week ]4 Page 7 of 9

b. PLDAT I TBT)

④AIB) -planB))=

PTBT=0.00672

0.07623=10.08827

Stat 302 - Fall 2020Week In Review

Problem 10. Many times, we see that the probability of a certain event depends on what eventshappened before. Let’s say Mrs. Fleming is the advisor of choir at a 7th-8th grade middle school.She wants to randomly select two students to attend a conference.

a. What is the probability that at least one student is in seventh grade?b. What is the probability that both students are in seventh grade?c. What is the probability that both students are in the same grade?

Shelby Cummings Week ]4 Page 8 of 9

7th : I 0•A . Plat least one student

-- 7th)

I - P(none 7th)= I- P(both 8th)

Both 8th : 1st→ 8thp --¥

2nd-pa 8th p = ÷plBoth 8th) = (EE) (YT)

= 0.2857

PCat least one-7th) = I-P(Both8th)

= I- O .2857=10.71437

Stat 302 - Fall 2020Week In Review

Problem 10. Many times, we see that the probability of a certain event depends on what eventshappened before. Let’s say Mrs. Fleming is the advisor of choir at a 7th-8th grade middle school.She wants to randomly select two students to attend a conference.

a. What is the probability that at least one student is in seventh grade?b. What is the probability that both students are in seventh grade?c. What is the probability that both students are in the same grade?

Shelby Cummings Week ]4 Page 8 of 9

7th : IO

b -Plboth 7th)

8€12BOTH 7th

1st-*7th p= to

2nd -p. 7th p = IT

PlBooth 7th)-- (1%2) (9/21)=10.19487

C. Plboth same grade)=

P(both 7th) t PLboth8th)

= 0.1948 t0.2857=10.48057

Stat 302 - Fall 2020Week In Review

Problem 11. A professor surveyed his students and asked them, on a typical weekend, howmany days do you spend studying. The probability distribution is shown below.

x 0 1 2 3P(X=x) 0.10 0.35 0.25 0.30

a. What is the probability that a randomly selected individual studied at least 2 days?b. What is the probability that a randomly selected individual studied no more than 2 days?c. What is the expected value?

Shelby Cummings Week ]4 Page 9 of 9

A. PIX ? 2) = PIX-2)t PCX

-- 3)

=0.25 t O .30=10.557b. PCXE 2) = plx=Dtpcx

--Dtp(X=2)

= O. I0 to .35 to .25=10.707

C. EX-op(X=x;) - sumof : possibilities itie,

ELX) = 10710.10) tCD 10.35) t(2)10.25

)t

(3) 10.30)

= 0 t 0.35to .SO to .90=11.757