Probability

1
Probability

• Randomness and Chance

2
Motivation

• We need concept of probability to make judgments
  about our hypotheses in the scientific method.
  Is the data consistent with our hypotheses?
• Example Suppose an old drug cures 50 percent of
  the time. Our new drug cures 6 out of 10
  patients. Is our new drug better?

3
Random Phenomenon

• A phenomenon is random if individual outcomes are
  uncertain, but there is a regular distribution of
  outcomes in many repetitions.
• Toss a coin.
• Toss a die.

4
Definition

• The probability of any outcome of a random
  phenomenon is the proportion of times the outcome
  would occur in a very long series of repetitions.
  Probability is a long-term frequency.
• Do simulation on web site.

5
Symmetry Definition

• Some random phenomenon occur under a situation of
  equally likely outcomes coin tossing, die
  tossing, lotteries, bingo, etc.
• The probability of rolling a 4 on a die is the
  ratio of the number of sides containing a 4
  divided by the total number of sides. On a fair
  die this means 1/6.

6
Sample Space

• Set of all possible outcomes is called the sample
  space, S.
• Coin toss SH, T
• Die toss S1, 2, 3, 4, 5, 6
• Toss coin twice SHH, TT, HT, TH

7
Event

• An event is a collection of outcomes.
• Die toss, event A2, 4, 6, even number of spots.

8
Probability

• Probability of an event is a number between zero
  and one, 0
• P(A) 1, means event has certainty.
• P(A) 0, means event is impossible.
• P(A-Complement) 1 P(A), that is, the
  probability of not A is simply 1 minus the chance
  of A happening.

9
Equally Likely Outcomes

• Probabilities under equally likely outcome case
  is simply the number of outcomes making up the
  event, divided by the number of outcomes in S.
• Example A die toss, A2, 4, 6, so
• P(A) 3/ 6 1/2 .5
• Example Coin toss, AH, P(A) 1/2 .5

10
Counting Problems

• How many arrangements of two digits, 0-9 ?
• Ans 1010 100
• How many phone numbers in an area code (7
  digits)?
• Ans 10 raised to 7th power, minus number of
  ineligible numbers, 10,000,000 a few.

11
Minnesota Daily 3

• Probability of winning the straight Daily 3?
• Ans equally likely outcomes, so answer is
  number of ways to win, divided by number of
  possible outcomes 1 / 1000 .0001

12
Powerball

• Grand Prize 1 in 120,526,770.00
• 100,000 1 in 2,939,677.32
• 5,000 1 in 502,194.88
• 100 1 in 12,248.66
• 100 1 in 10,685.00
• 7 1 in 260.61
• 7 1 in 696.85
• 4 1 in 123.88
• 3 1 in 70.39
• The overall odds of winning a prize are 1 in
  36.06.

13
Poker Hands

• There are 2,598,960 possible 5 card possibilities
  in poker, this is denominator for each hand.
• No pair approx 1/2.
• One pair 1,098,240 / 2,598,960 approx 1/2.
• Full house 3744 / den 1/694
• Royal Flush 4 / den 1 / 649,740
• Least likely hand wins pot.

14
Conditional Probability

• Consider two events A and B.
• What is the probability of A, given the
  information that B occurred? P(A B) ?
• Example Die toss, A1, 2, 3, B2, 4, 6.
• P(A B) P(A and B) / P(B) (1/6) / (3/6)
  1/3.

15
Conditional Probability

• Or better yet, calculate the chance of A
  happening out of the possible outcomes of B.
• The only way for A1, 2, 3 to occur out of
  B2, 4, 6, is a 2 outcome.
• P(A B) ways A could occur from B, divided
  by number of ways in B, 1/3.

16
Venn Diagram
17
Probability Problems

• P(Married) 59,920/103,870
• P(Married 18-29) 7842/ 22,512

18
Independence

• Events A and B are independent if
• P(AB) P(A), that is, if the additional
  knowledge of B does not change the probability of
  A happening.
• On the marriage problem, the events Marriage and
  18-29 are dependent because P(Marriage18-29)
  does not equal P(Marriage), so these events are
  dependent.
• In die tossing, event roll a 3 on roll 2 is
  independent of roll a 6 on roll1. P(3 6) 1/6
  P(3) 1/6.

19
Probability Trees
Heads
P(HH) 1/4
.5
.5
P(HT) 1/4
Heads
Tails
.5
.5
P(TH) 1/4
Heads
.5
Tails
.5
P(TT) 1/4
Tails
20
Two Coin Flips

• P(No Heads) P(TT) ¼
• P(One Head) P(HT) P(TH) 1/4 1/4 1/2
• P(Two Heads) P(HH) 1/4

21
Multiple Choice Exam
P(RRR) (1/4)(1/4)(1/4) .0156
Right
P(RRW) (1/4)(1/4)(3/4) .0468
Right
W
R
Right
W
W
Right
R
Wrong
W
Wrong
Right
P(WWW) (3/4)(3/4)(3/4).4218
Wrong
22
Multiple Choice Exam

• P(Right) 1/4 if guessing out of 4 choices.
• P(Wrong) 3/4 if guessing out of 4 choices.
• P(Three right) .0156, found by multiplying the
  branch probabilities.
• P(None right) .4218, found by multiplying the
  branch probabilities.
• P(One right) P(WWR) P(WRW) P(RWW)
  (3/4)(3/4)(1/4) same same 3(same).4218

23
Randomized Response Surveys

• Used to obtain truthful answers to sensitive
  questions like have you ever killed someone?
• Truthful answers not possible without some
  protection of subjects. Anonymity.
• Refer to page 362, problem 4.125, says have
  subject toss coin, if coin is heads, subject
  answers truthfully about plagiarizing paper. If
  toss is tails then subject answers Yes no matter
  what.

24
Randomized Response Survey
Yes
p
1-p
Heads
No
.5
.5
Answer is Yes
Tails
25
Randomized Response Survey

• Suppose we take a survey of 100 subjects and we
  get 61 yes answers. What is the probability
  someone plagiarized a paper? Symbol p on diagram.
• Probability tree result for answering yes should
  match the survey result of 61/100.
• From Tree P(yes) 1/2 1/2p.
• From Survey P(yes) approx 61/100 .61.
• Solve 1/2 1/2p .61 1/2p .61-.5
• 1/2p .11 p-hat 2.11 .22, so 22 percent
  of subjects have plagiarized a paper.

26
Solution Breakdown
Yes
11 yes responses
p
50
39 No
100
50 yes responses
Solution p 11/50 .22
27
Solution Breakdown

• Notice that we sacrifice the bottom 50 subjects
  in order to gain truthful answers from those that
  flipped a heads originally. Well worth the price
  in most cases.
• Notice that the investigator will not be able to
  identify a true yes response from a coin-flipped
  yes response.

28
Alternative Design
Yes
p
1-p
Heads
No
.5
.5
Heads
Yes
Tails
Tails
No
29
New Design

• Notice this design helps protect both the yes and
  no responses from being detected. In the old
  design, the no responses were clearly identified.
• Suppose there are 100 subjects and 48 yes
  responses, what is our estimate of p?
• This time solve (1/2)p 1/4 48/100,
• (1/2)p .23, p-hat 2.23 .46

30
New Design
23
48-2523
p
50
27
100 Subjects
25
50
25
Solution p 23/50 .46
31
HIV Testing

• Ch 4 problem 125. Given P(HIV) .9985,
  P(-HIV) .0015, P(noHIV).006,
  P(-noHIV).994.
• Assume one percent of population has the
  infection.

32
HIV Testing
.009985

.9985
.0015
Has HIV
-
.000015
.01
.99
.00594

.006
No HIV
.994
-
.98406
33
HIV Testing

• P( ) .009985 .00594 .015925
• P( HIV ) P(HIV and ) / P()
• .009985 / .015925 .627.
• This means that only 62.7 percent of those that
  test positive for HIV really have the disease.
  Hard to trust a positive blood test.

34
HIV Testing

• Compute P(No HIV -) . If a blood test comes
  back negative, can you trust the result?
• The reason we get these results is because we are
  screening for a relatively rare thing, HIV
  disease. Any time a rare disease is screened
  for, we get tremendous inefficiencies. Another
  example is mammography for breast cancer.
  Efficiency of screening improves as prevalence
  increases.

35
Screening Structure

Sensitivity
-
False Negatives
Prevalence
Population
False Positives

Specificity
-
36
Space Shuttle Challenger Disaster
Work
Work
Work
Work
.977
Work
.977
BOOM!!
Work
Fail
.977
P(Shuttle Launch) .9776 .869
Fail