MATH Midterm Solutions

Transcription

1 MATH Midtem Solutios Febuay A bag of mables cotais 4 which ae ed, 4 which ae blue ad 4 which ae gee. a How may mables must be chose fom the bag to guaatee that thee ae the same colou? We ca apply the pigeohole piciple whee the pigeos ae the mables ad the holes ae the colous. Sice thee ae 3 colous, to guaatee 3 of the same colou, we eed pigeos mables. b If you daw thee mables at adom fom the bag, what is the pobability that they ae the same colou? It makes o diffeece what colou is daw fist. Fom that poit, 3 mables left ae the same colou of the 11 emaiig so the pobability the secod is the same colou as the fist is 3. Fo the thid 11 mable, we must pick oe of the two emaiig fom the available 10 so this has pobability 2. Theefoe, the pobability of pickig all thee 10 the same colou is c How may mables must be chose fom the bag to guaatee that thee is at least oe of evey colou? This time, we ca see that it is possible to choose 8 whee oly two colous ae epeseted all ed ad blue fo example but ay 9 mables will cotai all colous of mables. Theefoe, the miimum equied is 9. d If you daw thee mables at adom fom the bag, what is the pobability that they ae all diffeet colous?

2 As befoe, the fist mable daw does ot matte. To choose oe that is of a diffeet colou is 8. The thee ae oly 4 mables left 11 of the colou ot yet epeseted, so the pobability of dawig oe of them is 4. Theefoe, the pobability of dawig mables of all thee 10 colous is Fo how may iteges fom 1 to 9999 is the sum of thei digits equal to 9? Fo all such iteges, we ecogize that this is the same as solvig the poblem of coutig the umbe of solutios to x 1 + x 2 + x 3 + x 4 9 whee x i 0. We could wite the digits x 1 though x 4 i ode to ceate a umbe betwee 1 ad Usig the esult fom class, we get that thee ae 220 solutios. Theefoe, thee ae positive iteges less tha which have this popety. 3. a Defie PA B. PA B PA B PB b Pove that if PA PB the PA B PB A PA B PA B PB PA B PA PB A c Defie what it meas fo A ad B to be idepedet evets. A ad B ae idepedet if they satisfy PA B PA, o equivaletly, PA B PA PB. Suppose thee ae thee supplies of compute pats X,Y ad Z. Whee 5% of X s poducts ae supeio quality, 10% of Y s poducts ae supeio quality ad 15% of Z s poducts ae supeio quality. A paticula stoe gets 50% of its pats fom X, 30% fom Y ad 20% fom Z.

3 d If a uit is puchased, what is the pobability that it is supeio quality? We will let S be the eve that a uit is of supeio quality. Sice the supeio poduct must have come fom oe of the thee supplies, we ca deduce that PS PS X + PS Y + PS Z e If a uit i the stoe is foud to be supeio quality, which supplie is most likely to have come fom? Fo each of the thee supplies, we calculate the appopiate coditioal pobabilities: PX S PX S PS PY S PZ S PY S PS PZ S PS So, the supeio quality uit is most likely fom eithe Y o Z each with the same pobability. 4. a Defie the mathematical expessio choose. b Pove Pascal s idetity: + 1!!! + 1

4 As see i class, thee ae at least 2 o 3 ways to pove this idetity. I will peset my favouite hee: Suppose thee ae + 1 people i a oom oe of whom is Bob. We wat to fid out how may ways thee ae to make a goup of of them. + 1 Of couse, this is equal to. We could also cout the umbe of ways as follows: If we make a goup without Bob, thee ae combiatios. If we iclude Bob, we must choose 1 moe fom the emaiig givig combiatios. Sice evey goup eithe 1 cotais Bob o it does t, we must have couted exactly the same umbe of combiatios. Theefoe, Let D {1, 2, 3,..., 52} ad defie s : D D by a Is s a well defied fuctio? 2x 1 if 1 x 26 sx 2x 52 if 27 x 52 We ca easily check that s is defied fo all x D ad that fo 1 x 26, 1 sx 51. Also, fo 27 x 52 we get that 2 sx 52. Theefoe, s is well defied. b Is s oe to oe? c Is s oto? If sx is odd the we ca calculate that x sx+1 ad if sx is eve 2 the we ca calculate that x sx+52. Sice this defies the ivese of 2 s, we have foud the ivese ad coclude that s is oe to oe ad oto.

5 d Fid a expessio fo s s. We simply apply the fuctio twice. Sice sx could be moe o less tha 26, we ae goig to get 4 cases istead of just two. s sx 4x 3 if 1 x 13 4x 54 if 14 x 26 4x 105 if 27 x 39 4x 156 if 40 x Let D {1, 2, 3,..., 52} ad let c be a costat such that 1 c 52. Defie f c : D D by x c if x > c f c x x + 52 c if x c a Pove that f c is a well defied fuctio. f c is defied fo all x. I each of the two cases 1 x c 52 whe x > c ad 1 x + 52 c 52 whe x c. b Pove that f c is oe to oe fo all c. As we did i questio 5, each case is a liea fuctio which will be oe to oe. But it could be the case that f c x 1 f c x 2 if x 1 > c ad x 2 c. But i this case we get that f c x 1 f c x 2 x 1 c x 2 c + 52 x 1 x This of couse is impossible sice the xs ae chose fom D which oly has values fom 1 to 52. Theefoe, it must be the case that f c is oe

6 to oe. c Pove that f c is oto fo all c. Sice f c is a map betwee sets of the same size ad f c is also oe to oe, we ca coclude that it must also be oto. d Fid a expessio fo f 1 c. fc 1 x x + c if x 52 c x 52 + c if x > 52 c To double check you could compute fc 1 idetity, though this is ot equied. f c to show that it is the