[SOLVED] CS AI EX A Observe a future annual high-water level of Lake Ontario.

EX A Observe a future annual high-water level of Lake Ontario.
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EX B Toss 4 coins (or a single coin 4 times).
EX C Deck of 6 cards labelled 1, 2, , 6. Choose two cards (dont worry about order) without replacement. (Note: If with replacement, order will necessarily become an issue.) Well let ij indicate the outcome that cards i and j are selected. (This is horrible but minimal notation.)
Sample Space
S = { ij : i = 1, 2, 3, 4, 5; j = (i + 1), , 6 }
= { 12, 13, 14, 15, 16, 23, 24, 25, 26, 34, 35, 36, 45, 46, 56 }
Events:
A = { 12, 13, 14, 23, 24, 34 } B = { 13, 24, 35, 46 }
C = { 15, 24 } D = { 15, 25, 35, 45 } E = { 14, 24, 34 }
Axioms of Probability
Probability is a function that assigns a number to an event. Only three rules are needed to get things off the ground in a fashion that agrees with our intuitive notion of what probability should be.
1. For any event A, 0 P(A) 1.
2. P(S) = 1.

3. For events A , A , that are mutually exclusive: P A = P(A ).
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Implication (way down the road): Imagine the experiment is repeated many times where the starting conditions are identical each time and the results of any previous experiments have no impact on the way future experiments unfold. The relative frequency of occurrence of an event will with virtual certainty approach the probability of that event. A probability can be thought of as an ideal relative frequency.
i i=1 i=1

Propositions
(2 and 4 do not appear in the text.) For events A and B.
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1) 2)
3)
4) 5)
P(A)+P(Ac)=1
P(A)= P(AB) P AB+ (This is event decomposition decomposing A into the mutually exclusive
()
c
parts occurring with and not with B.) If ABthen P(A)P(B)
P(AB)= P(AB) P(+ABc ) P(+AcB). (More decomposition.) P(AB)=P(A) P+(B) P(AB)
Note that Axiom 3 handles / simplifies 5 above for mutually exclusive events. It is simple to find probabilities of unions (at least one) for mutually exclusive events having known probabilities. Its also often easy to compute probabilities for intersections such as AB (coming attraction). Many problems involve unions of mutually exclusive intersections such as in the items 2 and 4 above.
Venn Diagrams are useful for two and three events (maybe more?).
Inclusion / Exclusion
Two events: P(AB)= P(A) P+(B) P(AB)
Three events: PABC PAPBPC PABPACPBC PABC
Four events:
PABCD PAPBPCPD
In general for a finite collection of events E1, E2, , En.
More succinctly.

PAB PAC PAD PBC PBD PCD
PABC PABD PACD PBCD
PABCD

c
Y1
/1
.. ,t –
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c..,,.,)<VxX X).._3..s;ts ~l’- ~(.. :;. (4uB0v ,ov-:.:Page 4,oc:… …u- (,’–“-’11( ( ~&. {c._.kl’0Page 5..I, ….SoPage 6 Page 7—- ,,to C~ ~ o ~ s …Page8 1 PC–Page9 I() . –IPage 10() Inclusion / ExclusionTwo events: P(AB)= P(A) P+(B) P(AB)Three events: PABC PAPBPC PABPACPBC PABCFour events:Page 12PABCD PAPBPCPD In general for a finite collection of events E1, E2, … , En.More succinctly. PAB PAC PAD PBC PBD PCD PABC PABD PACD PBCD PABCDPage 13EX CARDS There are n cards labeled 1, 2, …, n. The cards are randomly permuted into bins also labeled 1, 2, … , n. (one card per bin). Whats the probability that at least one cards label matches the bins label? Start by letting Ai denote the event that card i finds its way into bin i. the card matches the bin. P(Ai) = 1/n. However,nnn1 PA = P(A )= = n i=1 i=1 i=1i1 (the Ai are not mutually exclusive). (1 doesnt make sense as a probability for this. Its possible that all cards fail to match you can easily arrange a deck to force it.) Booles inequality is ai nndud too: PA P(A )=1. (The book treats this as n men, each with a hat. The hats are thrown into aii i=1 EX REJ From a lot of 25 items, 5 are randomly selected and tested. The lot is rejected if over 2 of the sampledi=1box, then each man reaches in a randomly pulls out one hat. How probable is it at least one man gets his hat?)items are defective. If 8 of the 25 items are indeed defective, what is the probability the lot is rejected?EX POK A 5-card poker hand, randomly drawn from a standard deck (13 cards A K in each of four suits . Find the probabilities of a) two pairs and b) a straight. Some poker knowledge is helpful. A straight is consecutive cards not all of the same suit. A is treated as both low (A-2-3-4-5) and high (10-J-Q-K-A).R for combinatorics.> choose(13,2)*choose(4,2)^2*choose(44,1)/choose(52,5) [1] 0.04753902
> (10*4^5-40)/choose(52,5)
[1] 0.003924647
> ((10*4^5-40)/choose(52,5))^-1
[1] 254.8
EX 27 An urn contains 7 black and 4 white balls, laid out in order (Spot 1, Spot 2, , Spot 11). If we scan the balls in that order, whats the probability we first see a white ball in Spot i? First: Identify possible values for i.
EX 28 A bucket (I hate urns; who talks about urns?) contains 5 black, 7 yellow and 9 red balls. If a set of 3 balls is randomly selected, a) what is the probability they are all of the same color? b) of different color? Repeat assuming that balls are replaced after being selected (sampling with replacement).
EX BD With n people in a room, how likely is it that two have the same birthday (month and day only). Assume all birthdays are equally likely (not too far from the truth) and ignore Feb 29.
> n = 30
> v <- seq(365,365-n+1) >v
[1] 365 364 363 362 361 360 359 358 357 356 355 354 353 352 351 [16] 350 349 348 347 346 345 344 343 342 341 340 339 338 337 336 > v_over_365 <- v/365> v_over_365
[1] 1.0000000 0.9972603 0.9945205 0.9917808 0.9890411 0.9863014
[7] 0.9835616 0.9808219 0.9780822 0.9753425 0.9726027 0.9698630 [13] 0.9671233 0.9643836 0.9616438 0.9589041 0.9561644 0.9534247 [19] 0.9506849 0.9479452 0.9452055 0.9424658 0.9397260 0.9369863 [25] 0.9342466 0.9315068 0.9287671 0.9260274 0.9232877 0.9205479 > prod(v_over_365) # prod = product
[1] 0.2936838
> n <- 50> v <- seq(365,365-n+1)> v_over_365 <- v/365> prod(v_over_365)
[1] 0.02962642
P(all different birthdays)

+L
, vJ(J.
(!_ # Everything after an > is a command (R prompts commands with > )
> # + indicates a line break; its not user-typed, its Rs way of outputting it > # # indicates comment
> results <- c()> n <- 13> s <- 0> f <- 1> j <- -1> for (i in 1:n) {
+ j = -1*j
+ f <- f*i+ s <- s + j/f++ results[i] <- s+}> error <- results – (1-exp(-1))*rep(1,n)> rel_pct_error <- abs(100*error/results)> round_results <- formatC(results,digits=10,format=”f”)> round_rel_pct_error <- formatC(rel_pct_error,digits=8,format=”f”)> # approximate result
> formatC(1-exp(-1),digits=10)
[1] 0.6321205588
> data.frame((1:n),round_results,round_rel_pct_error)
With two cards they are either in order or not. So there are 0 matches or 2. The probability of at least one match is 0.5. The approximate (large n) probability is 0.632. The approximation is 26.4% high.
There are six orderings of three cards into bins 1, 2, 3: 123, 132, 213, 231, 312, 321. Four of them include a match: the probability of at least one match is 2/3 = 0.667. The approximation is 5.2% low.
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X.1.n. round_results round_rel_pct_error
11.0000000000
20.5000000000
30.6666666667
40.6250000000
50.6333333333
60.6319444444
70.6321428571
80.6321180556
90.6321208113
100.6321205357
110.6321205608
120.6321205587
130.6321205588
36.78794412
26.42411177
5.18191618
1.13928941
0.19149071
0.02786865
0.00352742
0.00039601
0.00003994
0.00000366
0.00000031
0.00000002
0.00000000
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