The document discusses probability and statistical inference. It provides definitions and examples of key concepts in probability such as experiments, sample spaces, events, and rules for calculating probabilities of events. Examples include calculating the probability of getting a 7 when rolling two dice and the probability of testing positive for a disease given the accuracy of the test. The document also provides examples of applying probability concepts to problems involving cards and birthdays.

1. StatiSticS
Part-ii
Mr. MuteeullahMr. Muteeullah
ChannaChanna

2.  It is a discipline that allows us toIt is a discipline that allows us to
estimate unknown quantities byestimate unknown quantities by
making some elementarymaking some elementary
measurements.measurements.
 Using these estimates we can thenUsing these estimates we can then
 make Predictions and Forecast themake Predictions and Forecast the
FutureFuture
inferential
StatiSticS?

3. ““Statistical inference is an art of makingStatistical inference is an art of making
conclusion about any statisticalconclusion about any statistical
characteristic of the populationcharacteristic of the population
(parameter) using sample information(parameter) using sample information
(limited information)”.(limited information)”. In other wordsIn other words
““The process of drawing inferencesThe process of drawing inferences
about a population on the basis ofabout a population on the basis of
information contained in a sampleinformation contained in a sample
taken from the population is calledtaken from the population is called
Statistical Inferences.”Statistical Inferences.”
Statistical inferences usually classifiedStatistical inferences usually classified
in two parts:in two parts:
StatiStical
inferenceS

4. 1.1. ““Estimation is the first part of InferentialEstimation is the first part of Inferential
Statistics. Estimation is a process ofStatistics. Estimation is a process of
developing single value or a class ofdeveloping single value or a class of
values as an estimate or class of estimatesvalues as an estimate or class of estimates
of the understudy parameter usingof the understudy parameter using
sample values Xsample values X11, X, X22 ,…., X,…., Xnn from thefrom the
population.”population.”
2.2. ““Testing of hypothesis is the second mainTesting of hypothesis is the second main
and major part of Inferential Statistics.and major part of Inferential Statistics.
The procedure which enables us to decideThe procedure which enables us to decide
on bases of information obtain from theon bases of information obtain from the
sample data, whether to accept or rejectsample data, whether to accept or reject
statement or assumption about the valuestatement or assumption about the value
of population parameter. Such statementof population parameter. Such statement
which may or may not be true is calledwhich may or may not be true is called
statistical hypothesis.statistical hypothesis.
1. eStimation
2. teSting of hyPotheSiS

5. ““Any statement or assumptionAny statement or assumption
regarding any statistical characteristicregarding any statistical characteristic
of the population is called statisticalof the population is called statistical
hypothesis.”hypothesis.”
StatiStical hyPotheSiS
SimPle and comPoSite
hyPotheSiS
““Any statistical hypothesis is said to beAny statistical hypothesis is said to be
a simple hypothesis if it is expressed bya simple hypothesis if it is expressed by
single numerical value.” wherelse,single numerical value.” wherelse,
““Any statistical hypothesis is said to beAny statistical hypothesis is said to be
composite hypothesis if a class ofcomposite hypothesis if a class of
numerical values expresses it.numerical values expresses it.

6. ““The null hypothesis is the hypothesisThe null hypothesis is the hypothesis
that is always tested for possiblethat is always tested for possible
rejection or nullification under therejection or nullification under the
assumption that is true, it is denoted byassumption that is true, it is denoted by
HHoo”.”.
““The alternative hypothesis is usuallyThe alternative hypothesis is usually
negation of the null hypothesis,negation of the null hypothesis,
represent the conclusion that would berepresent the conclusion that would be
drawn if evidence of “guilt” were found,drawn if evidence of “guilt” were found,
it is denoted by Hit is denoted by H11”.”.
null and alternative
hyPotheSiS

7.  It is a discipline that allows us toIt is a discipline that allows us to
estimate unknown quantities byestimate unknown quantities by
making some elementarymaking some elementary
measurements.measurements.
Using these estimates we can thenUsing these estimates we can then
make Predictions and Forecast themake Predictions and Forecast the
FutureFuture
What iS
inferential
StatiSticS?

8. chaPter 1
Probability

9. Can you make money playing theCan you make money playing the
Lottery?Lottery?
Let us calculate chances of winning.Let us calculate chances of winning.
To do this we need to learn some basicTo do this we need to learn some basic
rules about probability.rules about probability.
These rules are mainly just ways ofThese rules are mainly just ways of
formalising basic common sense .formalising basic common sense .
Example: What are the chances thatExample: What are the chances that
you get a HEAD when you toss a coin?you get a HEAD when you toss a coin?
Example: What are the chances youExample: What are the chances you
get a combined total of 7 when you rollget a combined total of 7 when you roll
two dice?two dice?
conSider
a real Problem

10. AnAn EExperimentxperiment leads to a singleleads to a single
outcome which cannot be predictedoutcome which cannot be predicted
with certainty.with certainty.
Examples-Examples-
Toss a coinToss a coin:: head or tailhead or tail
Roll a dieRoll a die:: 1, 2, 3, 4, 5, 61, 2, 3, 4, 5, 6
Take medicineTake medicine:: worse, same, betterworse, same, better
Set of allSet of all outcomesoutcomes -- SSampleample SSpacepace..
Toss a coinToss a coin Sample space = {h,t}Sample space = {h,t}
Roll a dieRoll a die Sample space = {1, 2, 3, 4, 5, 6} Sample space = {1, 2, 3, 4, 5, 6}
1.1 exPerimentS

11. TheThe PProbabilityrobability of aof an outcome is an outcome is a
number between 0 and 1number between 0 and 1 thatthat
measures themeasures the likelihood that thelikelihood that the
outcome will occuroutcome will occur when thewhen the
experiment is performed.experiment is performed.
(0=impossible, 1=certain).(0=impossible, 1=certain).
Probabilities of all sample pointsProbabilities of all sample points
must sum to 1.must sum to 1.
Long run relative frequencyLong run relative frequency
interpretation.interpretation.
EXAMPLE:EXAMPLE: Coin tossing experimentCoin tossing experiment
P(H)=0.5 P(T)=0.5P(H)=0.5 P(T)=0.5
1.2 Probability

12. AnAn eventevent is a specific collection ofis a specific collection of
sample points.sample points.
The probability of an event A isThe probability of an event A is
calculated by summing thecalculated by summing the
probabilities of theprobabilities of the outcomesoutcomes in thein the
sample space for A.sample space for A.
1.3 eventS

13. Define the experiment.Define the experiment.
List the sample points.List the sample points.
Assign probabilities to the sampleAssign probabilities to the sample
points.points.
Determine the collection of sampleDetermine the collection of sample
points contained in the event of interest.points contained in the event of interest.
Sum the sample point probabilities toSum the sample point probabilities to
get the event probability.get the event probability.
1.4 StePS for
calculating
ProbailitieS

14. examPle:
the game of craPS
In Craps one rolls two fair dice.In Craps one rolls two fair dice.
What is the probability of theWhat is the probability of the
sum of the two dice showing 7?sum of the two dice showing 7?

15. (1,6)
(2,5)
(3,4)
(4,3)
(5,2)
(6,1)
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

16. 1.5 Equally likEly
outcomEs
So the Probability of 7 whenSo the Probability of 7 when
rolling two dice is 1/6rolling two dice is 1/6
This example illustrates theThis example illustrates the
following rule:following rule:
In a Sample Space S of equallyIn a Sample Space S of equally
likely outcomes. Thelikely outcomes. The
probability of the event A isprobability of the event A is
given bygiven by
P(A) = #A / #SP(A) = #A / #S
That is the number of outcomesThat is the number of outcomes
in A divided by the total numberin A divided by the total number
of events in S.of events in S.

17. AA compound eventcompound event is a composition ofis a composition of
two or more other events.two or more other events.
AACC
:: TheThe ComplementComplement of A is the eventof A is the event
thatthat A does not occurA does not occur
AA∪∪BB :: TheThe UUnionnion of two events A andof two events A and
B is the event that occurs ifB is the event that occurs if either A oreither A or
B or both occurB or both occur, it consists of all sample, it consists of all sample
points that belong to A or B or both.points that belong to A or B or both.
AA∩∩BB:: TheThe IIntersectionntersection of two eventsof two events
A and B is the event that occurs ifA and B is the event that occurs if bothboth
A and B occurA and B occur, it consists of all sample, it consists of all sample
points that belong to both A and Bpoints that belong to both A and B
1.6 sEts

18. 1.7 Basic
ProBaBility rulEs
P(AP(Acc
)=1-P(A))=1-P(A)
P(P(AA∪∪BB)=P(A)+P(B)-P()=P(A)+P(B)-P(AA∩∩BB))
Mutually Exclusive Events areMutually Exclusive Events are
events which cannot occur atevents which cannot occur at
the same time.the same time.
PP(AA∩∩BB)=0)=0 for Mutuallyfor Mutually
Exclusive Events.Exclusive Events.

19. 1.8 conditional
ProBaBility
P(A | B) ~ Probability of AP(A | B) ~ Probability of A
occuring given that B hasoccuring given that B has
occurred.occurred.
P(A | B) =P(A | B) = PP(AA∩∩BB) / P(B)) / P(B)
Multiplicative Rule:Multiplicative Rule:
PP(AA∩∩BB))
= P(A|B)P(B)= P(A|B)P(B)
= P(B|A)P(A)= P(B|A)P(A)

20. 1.9 indEPEndEnt
EvEnts
A and B are independent eventsA and B are independent events
if the occurrence of one eventif the occurrence of one event
does not affect the probabilitydoes not affect the probability
of the othe event.of the othe event.
If A and B are independent thenIf A and B are independent then
P(A|B)=P(A)P(A|B)=P(A)
P(B|A)=P(B)P(B|A)=P(B)
PP(AA∩∩BB)=P(A)P(B))=P(A)P(B)

21. chaPtEr 1
ProBaBility
EXamPlEs

22. Probability as
a matter of
life and death

23. Positive Test for Disease
1 in every 10000 people in Ireland1 in every 10000 people in Ireland
suffer from AIDSsuffer from AIDS
There is a test for HIV/AIDSThere is a test for HIV/AIDS
which is 95% accurate.which is 95% accurate.
You are not feeling well and youYou are not feeling well and you
go to hospital where yourgo to hospital where your
Physician tests you.Physician tests you.
He says you are positive for AIDSHe says you are positive for AIDS
and tells you that you have 18and tells you that you have 18
months to live.months to live.
How should you react?How should you react?

24. Positive Test for Disease
 Let D be the event that youLet D be the event that you
have AIDShave AIDS
 Let T be the event that you testLet T be the event that you test
positive for AIDSpositive for AIDS
 P(D)=0.0001P(D)=0.0001
 P(T|D)=0.95P(T|D)=0.95
 P(D|T)=?P(D|T)=?

25. Positive Test for Disease
)9999.0)(05.0()0001.0)(95.0(
)0001.0)(95.0(
)()|()()|(
)()|(
)()(
)()|(
}){}({
)()|(
)(
)(
)|(
+
=
+
=
+
=
=
=
CC
C
C
DPDTPDPDTP
DPDTP
DTPDTP
DPDTP
DTDTP
DPDTP
TP
TDP
TDP



001897.0=

26. chaPtEr 1
EXamPlEs
Example 1.1Example 1.1
S={A,B,C}S={A,B,C}
P(A) = ½P(A) = ½
P(B) = 1/3P(B) = 1/3
P(C) = 1/6P(C) = 1/6
What is P({A,B})?What is P({A,B})?
What is P({A,B,C})?What is P({A,B,C})?
List all events Q such thatList all events Q such that
P(Q) = ½.P(Q) = ½.

27. chaPtEr 1
EXamPlEs
Example 1.2Example 1.2
Suppose that a lecturer arrivesSuppose that a lecturer arrives
late to class 10% of the time,late to class 10% of the time,
leaves early 20% of the timeleaves early 20% of the time
and both arrives late ANDand both arrives late AND
leaves early 5% of the time.leaves early 5% of the time.
On a given day what is theOn a given day what is the
probability that on a given dayprobability that on a given day
that lecturer will either arrivethat lecturer will either arrive
late or leave early?late or leave early?

28. chaPtEr 1
EXamPlEs
Example 1.3Example 1.3
Suppose you are dealt 5 cardsSuppose you are dealt 5 cards
from a deck of 52 playing cards.from a deck of 52 playing cards.
Find the probability of theFind the probability of the
following eventsfollowing events
1. All four aces and the king of
spades
2. All 5 cards are spades
3. All 5 cards are different
4. A Full House (3 same, 2
same)

29. Chapter 1
examples
Example 1.4Example 1.4
The Birthday ProblemThe Birthday Problem
Suppose there are N people in aSuppose there are N people in a
room.room.
How large should N be so thatHow large should N be so that
there is a more than 50% chancethere is a more than 50% chance
that at least two people in thethat at least two people in the
room have the same birthday?room have the same birthday?

30. Number in Room Prob at least 2 have same birthday
1 0.00
2 0.00
3 0.01
4 0.02
5 0.03
6 0.04
7 0.06
8 0.07
9 0.09
10 0.12
11 0.14
12 0.17
13 0.19
14 0.22
15 0.25
16 0.28
17 0.32
18 0.35
19 0.38
20 0.41
21 0.44
22 0.48
23 0.51
24 0.54
25 0.57
26 0.60
27 0.63
28 0.65
29 0.68
30 0.71
31 0.73
32 0.75
33 0.77
34 0.80
35 0.81
36 0.83
37 0.85
38 0.86
39 0.88
40 0.89
41 0.90
42 0.91
43 0.92
44 0.93
45 0.94
46 0.95
47 0.95
48 0.96
49 0.97
50 0.97
51 0.97
52 0.98
53 0.98
54 0.98
55 0.99
56 0.99
57 0.99

31. Chapter 1
examples
Example 1.4Example 1.4
Children are born equally likelyChildren are born equally likely
as Boys or Girlsas Boys or Girls
My brother has two childrenMy brother has two children
(not twins)(not twins)
One of his children is a boyOne of his children is a boy
named Lukenamed Luke
What is the probability that hisWhat is the probability that his
other child is a girl?other child is a girl?

32. Example 1.5
The Monty Hall Problem
 Game ShowGame Show
 3 doors3 doors
 1 Car & 2 Goats1 Car & 2 Goats
 You pick a door - e.g. #1You pick a door - e.g. #1
 Host knows what’s behind allHost knows what’s behind all
the doors and he opens anotherthe doors and he opens another
door, say #3, and shows you adoor, say #3, and shows you a
goatgoat
 He then asks if you want toHe then asks if you want to
stick with your original choicestick with your original choice
#1, or change to door #2?#1, or change to door #2?

33. Ask Marilyn.
Parade Magazine Sept 9 1990
 Marilyn vos SavantMarilyn vos Savant
 Guinness Book of RecordsGuinness Book of Records
-Highest IQ-Highest IQ
 ““Yes you should switch. TheYes you should switch. The
first door has a 1/3 chance offirst door has a 1/3 chance of
winning while the second has awinning while the second has a
2/3 chance of winning.”2/3 chance of winning.”
 Ph.D.s - Now two doors, 1 goatPh.D.s - Now two doors, 1 goat
& 1 car so chances of winning& 1 car so chances of winning
are 1/2 for door #1 and 1/2 forare 1/2 for door #1 and 1/2 for
door #2.door #2.
 ““You are the goat”You are the goat” - Western- Western
State University.State University.

34. Who’s right?
 At the start, the sample space is:At the start, the sample space is:
{{CCGG,GG, GGCG,CG, GGGCGC}}
 Pick a door e.g. #1Pick a door e.g. #1
 1 in 3 chance of winning1 in 3 chance of winning
 Host shows you a goat so nowHost shows you a goat so now
{{CCGGGG,, GGCCGG,, GGGGCC}}
 So Marilyn was right, you shouldSo Marilyn was right, you should
switch.switch.

35. Not convinced?
 Imagine a game with 100 doors.Imagine a game with 100 doors.
 1 F430 Ferrari, 99 Goats.1 F430 Ferrari, 99 Goats.
 You pick a door.You pick a door.
 Host opens 98 of the 99 otherHost opens 98 of the 99 other
doors.doors.
 Do you stick with your originalDo you stick with your original
choice?choice? Prob = 1/100Prob = 1/100
 Or move to the unopened door.Or move to the unopened door.
Prob = 99/100Prob = 99/100

36. Boys, Girls
and Monty Hall
 Sample Space ( listing oldest childSample Space ( listing oldest child
first)first)
 {GG, BG, GB, BB}{GG, BG, GB, BB}
 Equally likely eventsEqually likely events
 One child is a boy:One child is a boy:
 GG is impossibleGG is impossible
 {BG, GB, BB} =>{BG, GB, BB} =>
 P(OC = G) = 2/3P(OC = G) = 2/3
 Luke is 6 months old.Luke is 6 months old.
 {GB, BB} =>{GB, BB} => P(OC = G) = 1/2P(OC = G) = 1/2

37. Odd sOCks
It is winter and the ESB are onIt is winter and the ESB are on
strike. This morning when youstrike. This morning when you
woke up it was dark. In your sockwoke up it was dark. In your sock
drawer there was one pair of twodrawer there was one pair of two
black socks and one odd brownblack socks and one odd brown
one.one.
Are you more or less likely to beAre you more or less likely to be
wearing matching socks today?wearing matching socks today?

38. exams
CampusCampus FemaleFemale
Pass RatePass Rate
MaleMale
Pass RatePass Rate
BelfieldBelfield 40%40% 33%33%
ET/ET/
CarysfortCarysfort
etc.etc.
75%75% 71%71%
Seeing this evidence amaleSeeing this evidence amale
student takes UCD to courtstudent takes UCD to court
saying there is disciminationsaying there is discimination
against male students. UCDagainst male students. UCD
gathers all it’s exam informationgathers all it’s exam information
together and reports thetogether and reports the
following.following.

39. exam
pass rates
Overall Female pass rate is 56%Overall Female pass rate is 56%
Overall Male pass rate is 60%Overall Male pass rate is 60%
HOW CAN THIS BE?HOW CAN THIS BE?
Clearly UCDClearly UCD areare LYING !LYING !
CampusCampus FemaleFemale
Pass RatePass Rate
MaleMale
Pass RatePass Rate
BelfieldBelfield 40%40% 33%33%
ET/ET/
CarysfortCarysfort
etc.etc.
75%75% 71%71%

40. simpsOn’s
paradOx
Overall Female pass rate is 56%Overall Female pass rate is 56%
Overall Male pass rate is 60%Overall Male pass rate is 60%
CampusCampus FemaleFemale
Pass RatePass Rate
MaleMale
Pass RatePass Rate
BelfieldBelfield 40%40%
= 20/50= 20/50
33%33%
= 10/30= 10/30
ET/ET/
CarysfortCarysfort
etc.etc.
30/4030/40
=75%=75%
50/7050/70
= 71%= 71%
50/9050/90
= 56%= 56%
60/10060/100
=60%=60%

41. hit and rUn
Once upon a time in Hicksville,Once upon a time in Hicksville,
USA there was a night-time hit andUSA there was a night-time hit and
run accident involving a taxi. Thererun accident involving a taxi. There
are two taxi companies inare two taxi companies in
Hicksville, Green and Blue. 85%Hicksville, Green and Blue. 85%
of taxis are Green and 15% areof taxis are Green and 15% are
Blue. A witness identified the taxiBlue. A witness identified the taxi
as being Blue. In the subsequentas being Blue. In the subsequent
court case the judge ordered thatcourt case the judge ordered that
the witness’s observation under thethe witness’s observation under the
conditions that prevailed that nightconditions that prevailed that night
be tested. The witness correctlybe tested. The witness correctly
identified each colour of taxi 80%identified each colour of taxi 80%
of the time.of the time.

42. hit and rUn
What is the probability that it wasWhat is the probability that it was
indeed a blue taxi that was involvedindeed a blue taxi that was involved
in the accident?in the accident?

43. DNA
You are holiday in Belfast and anYou are holiday in Belfast and an
explosion destroys the Odesseyexplosion destroys the Odessey
arena.arena.
You are seen running from theYou are seen running from the
explosion and are arrested.explosion and are arrested.
You are subsequently charged withYou are subsequently charged with
being a member of a prescribedbeing a member of a prescribed
paramilitary organisation and withparamilitary organisation and with
causing the explosion.causing the explosion.
In court you protest your innocence.In court you protest your innocence.
However the PSNI have DNAHowever the PSNI have DNA
evidence they claim links you to theevidence they claim links you to the
crime.crime.

44. DNA
Their forensic scientist delivers theTheir forensic scientist delivers the
following vital evidence.following vital evidence.
The forensic scientist indicates thatThe forensic scientist indicates that
DNA found on the bomb matchesDNA found on the bomb matches
your DNA.your DNA.
Your lawyer at first disputes thisYour lawyer at first disputes this
evidence and hires an independentevidence and hires an independent
scientist.scientist.
However the second forensicHowever the second forensic
scientist also says that the DNAscientist also says that the DNA
matches yours and that there is a 1 inmatches yours and that there is a 1 in
500 million probability of the500 million probability of the
match.match.

45. DNA
What do you do?What do you do?
It appears as if you are going toIt appears as if you are going to
spend the rest of your days in jail.spend the rest of your days in jail.

46. The NATioNAl
loTTery

47. “I lied, cheated and stole to
become a millionaire. Now
anybody at all can win the
lottery and become a
millionaire”

48. GAME #1: LOTTO 6/42
 What are the chance of winningWhat are the chance of winning
with one selection of 6with one selection of 6
numbers?numbers?
MatchesMatches Chances of WinningChances of Winning
66 1 in 5,245,7861 in 5,245,786
55 1 in 24,2861 in 24,286
44 1 in 5551 in 555

49. GAME #1: LOTTO 6/42
Expected WinningsExpected Winnings
Only consider JackpotOnly consider Jackpot
1 Euro get 1 play1 Euro get 1 play
E(win)= Jackpot*(1/5,245,786) –E(win)= Jackpot*(1/5,245,786) –
1Euro*(5,245,785/5,245,786)1Euro*(5,245,785/5,245,786)
E(win)=E(win)=
Jackpot*Jackpot*0.0000001910.000000191--
0.9999998090.999999809
If only one jackpot winner then:If only one jackpot winner then:
Positive E(win) ifPositive E(win) if
Jackpot >5,245,785Jackpot >5,245,785

50. LOTTO 6/42
 The average time to win each of the prizes isThe average time to win each of the prizes is
given by:given by:
 Match 3 with BonusMatch 3 with Bonus 2 Years, 6 Weeks2 Years, 6 Weeks
 Match 4Match 4 2 Years, 8 Months2 Years, 8 Months
 Match 5Match 5 116 Years, 9 Months116 Years, 9 Months
 Match 5 with BonusMatch 5 with Bonus 4323 Years, 5 Months4323 Years, 5 Months
 Share in JackpotShare in Jackpot 25,220 Years25,220 Years

51. TossiNg A fAir coiN

52. TossiNg A coiN!
 You are joking!You are joking!
 That is boring … no question about it!That is boring … no question about it!
 1957 Second edition of William Feller’s1957 Second edition of William Feller’s
Textbook includes a chapter on coin-Textbook includes a chapter on coin-
tossing.tossing.
 Introduction:Introduction: “The results concerning“The results concerning
…coin-tossing show that widely held…coin-tossing show that widely held
beliefs … are fallacious. These resultsbeliefs … are fallacious. These results
are so amazing and so at variance withare so amazing and so at variance with
common intuition that evencommon intuition that even
sophisticated colleagues doubted thatsophisticated colleagues doubted that
coins actually misbehave as theorycoins actually misbehave as theory
predicts.”predicts.”

53. TossiNg A coiN!
 Toss a coin 2N times.Toss a coin 2N times.
 Law of Averages:Law of Averages:
 As N increases the chances thatAs N increases the chances that
there are equal numbers of headsthere are equal numbers of heads
and tails among the 2N tossesand tails among the 2N tosses
increases.increases.
 LimLim N->N->∞∞ P( #H = #T ) = 1.P( #H = #T ) = 1.
 In the limit as N tends to infinityIn the limit as N tends to infinity
the probability of matchingthe probability of matching
numbers of heads and tailsnumbers of heads and tails
approaches 1.approaches 1.

54. roseNcrANTz
AND
guilDeNsTerN
Are DeAD

55. Prob of equAl
Numbers of h AND T
# of# of
tossestosses
22 44 66 88 1010
½½ 3/83/8 5/165/16 35/12835/128 63/25663/256
ProbProb 0.50.5 0.3750.375 0.31250.3125 0.2730.273 0.2460.246