Math 151 Homework 2 Solutions (Winter 2015)
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1 Math 5 Homewo 2 Solutions (Winte 25 Polem 3. (a Let A and A 2 e the events that the fist and the second selected alls, espectively, ae white. Also let B and B 2 e the events that the thid and the fouth selected alls, espectively, ae lac. Then y multiplication ule P (A A 2 B B 2 P (A P (A 2 A P (B A A 2 P (B 2 A A 2 B ( One can easily see that the poaility of the event that the i th and j th alls ae lac and the est ae white out of the fist selections does not depend on i and j. That s ecause usin multiplication ule this poaility can e witten as a poduct of factions which has ({2,, 6, 8} as denominatos and {5, 7, 7, 9} as numeatos in some ode. Since thee ae 2 6 ways to choose the position of lac alls amon the fist selections the desied 35 poaility is equal to Polem 3.23 Let A i w e the event that the all chosen fom i th un is white. Similaly let A i e the event that the all chosen fom i th un is ed. Then (a P (A 2 w P (A 2 w A wp (A w + P (A 2 w A P (A ( P (A w A 2 w P (A wa 2 w P (A 2 w Polem 3.28 P (A2 w A wp (A w P (A 2 w Let A e the event that the fist ace is the 2th cad to appea. Also let B and C e the events that 2st cad is the ace of spades and two of clus, espectively. (a If the fist ace appeas in the 2th cad then the fist 9 cads must e chosen fom the est of 8 cad in ( 8 9 ways and each such 9 cad can appea in 9! diffeent odes. Since the fist ace can e chosen in ( ( ways thee ae exactly ( 8 9 9! outcomes in A. Hence ( 8 P (A ( 9 9!. Similaly to compute the nume of outcomes in AB note that the only 2! (
2 diffeence is that the ace in 2th position now can e chosen in ( 3 ways. Theefoe thee ae ( 3 8 ( ( ( ! 9! outcomes in AB and P (AB 9. So P (B A P (AB 3 2! P (A 28. ( To compute the nume of outcomes in AC note that now the fist 9 cads must e chosen fom 7 cads and ace in the 2th position can e chosen in ( ( ways. So thee ae ( 7 ( 9 9! 7 outcomes in AC and P (AC ( 9 9!. Hence P (B A P (AB 29 2! P (A 536. Polem 3. ( 52 2 Let A e the event that the chosen cad is an ace and let B e the event that the chosen cad is in fact the cad that was intechaned. Since thee ae 27 cads afte the intechane P (B 27 and P (B c Oviously P (A B. Now oseve that conditional on the event Bc, any cad fom the emainin 5 cads has the same poaility to e dawn fom the second half. Since thee ae 3 aces amon that 5 cads we et P (A B c 3 5. So ( 52 2 P (A P (A BP (B + P (A B c P (B c Polem 3. Denote the othe pisones y B and C and suppose the jaile tells A that B will e set fee. Let X e the event that A will e executed and let Y e the event that B will e set fee. Also let Y and Z e the events that B and espectively C will e executed. Then P (X Y P (XY P (Y P (Y XP (X P (Y XP (X + P (Y Y P (Y + P (Y ZP (Z So the jaile was not iht i.e., the poaility of A ein executed does not chane Polem 3.6 (a Since Smith s siste has lue eyes oth of his paents must have at least one lue-eyed enes. Since paents also have own eyes they also must have at least one own-eyed enes. So they have one lue-eyed and one own-eyed enes. If A is the event that Smith has own eyes and B is the event that Smith possesses lue-eyed ene then ased on the infomation that oth paents have one own-eyed and one lue-eyed enes P (A 3 and P (AB 2. Thus P (B A P (AB P (A
3 ( Let C e the event that Smith s child will have lue eyes and let W e the event that Smith s wife has lue eyes. Then P (C W A P (C W BAP (B W A + P (C W B c AP (B c W A (c Let B and B 2 e the event that thei fist and second child, espectively, have own eyes. Then P (B 2 B W A P (B 2B W A P (B W A By Bayes s fomula P (B 2B W AP (W A P (B W AP (W A P (B 2B W A P (B W A P (B 2 B W A P (B 2 B W ABP (B W A + P (B 2 B W AB c P (B c W A and P (B W A P (B W ABP (B W A + P (B W AB c P (B c W A So P (B 2 B W A 3. Polem 3.67 (a Let A e the event that 2-out-of- system does not function. Fo evey outcome in A thee can e at most one functionin component. Hence P (A ( p i + i p i ( p j and poaility that the system functions is equal to P (A. ( Similaly if A is the event that 2-out-of- system does not function then P (A ( p i + i i j i p i ( p j + i j i i<j and poaility that the system functions is equal to P (A. p i p j i, j ( p (c If A i is the event that exactly i components function then P (A i ( n i p i ( p n i. So if F is the event that -out-of-n system functions then P (F n P (A i i n i ( n p i ( p n i i 3
4 Polem 3.7 Let X e the event that A olls the sum 9 in th oll and let Y e the event that B olls the sum 6 in th oll. Then P (X 36 and P (Y Now let W e the event that the ame stops in th oll and let W e the event that the final oll is made y A. Then ( P (W 2+ P Xi c Yi c X + P (X + P (Xi c P (Yi c ( i i ( ( 32 P (W A olls fist P W 2+ P (W Polem 3.9 Let A i e the event that jude i votes uilty and let X e the event that the defendant is in fact uilty. Then (a ( (c P (A 3 A A 2 P (A A 2 A 3 P (A A 2 P (A A 2 A 3 XP (X + P (A A 2 A 3 X c P (X c P (A A 2 XP (X + P (A A 2 X c P (X c ( ( ( ( P (A 3 A A c 2 A c A 2 P (A A c 2A 3 A c A 2 A 3 P (A A c 2 Ac A 2 2P (A A c 2A 3 2P (A A c 2 P (A A c 2A 3 P (A A c 2 P (A A c 2A 3 + P (A c A 2 A 3 P (A A c 2 + P (Ac A 2 P (A A c 2A 3 XP (X + P (A A c 2A 3 X c P (X c P (A A c 2 XP (X + P (A A c 2 Xc P (X c ( ( P (A 3 A c A c 2 P (Ac A c 2A 3 P (A c Ac 2 P (Ac A c 2A 3 XP (X + P (A c A c 2A 3 X c P (X c P (A c Ac 2 XP (X + P (Ac Ac 2 Xc P (X c.7 ( ( ( ( (y symmety
5 Polem 3.T.7 (a Conside a new expeiment whee all the alls ae emoved fom the un one y one. Then in the fist expeiment the event that all the emainin alls ae white is now equivalent to the event that the last emoved all is white. Let S e the all possile ways of tain all alls fom the un i.e. S is the sample space. To count the outcomes in S it is enouh to conside white alls and when they wee withdawn. Since we can put n white alls in n + m possile withdawn position in ( ( n+m n diffeent ways we et S n+m n. Now let A e the event that the last emoved all is white. To count the nume of outcomes in A we just need to conside emainin n white alls and thei withdawal positions. So A ( n+m n. Theefoe P (A A S ( n+m n ( n+m n n n + m ( We can solve this in two diffeent ways. Fist solution: Aane names of the fish on a line as you emove them fom the pond. Then evey emoval of all fish fom the pond coesponds to a stin of lenth + + with Reds, Bluess and Geens. Thus the sample space has ( ++,, outcomes in it. The event R that the Red fish extinct fist coesponds to the all aanements whee last Red appeas efoe the last Blue and the last Geen. Now let R e the collection of all aanements whee the last Red comes efoe the last Blue and the last fish was emoved is Geen. Also let R e the collection of all aanements whee the last Red comes efoe the last Geen and the last fish was emoved is Blue. Then R R R. To count the nume of outcomes in R let e the position of last Blue on the line. Then all Reds and Blues must e in the fist positions. Thus thee ae ( ( ( + ( + possile such aanements. So ( + R ++ + ( + ( ( whee we used the fomula ( ( m m + m+ ( m + + m+ ( m m++ m+. Similaly, R ( ( Since R R + R we deduce that P (R ( + ( ++ + ( + + ( ( ++,, Second solution: Aain let R e the event that Red fish is the fist species to ecome extinct. Let G and B e the events that Geen fish, espectively Blue fish, is the last species to ecome extinct. Then usin pat (a we et P (B and P (G. Also oseve that conditionin on the event B the event R has the same poaility as if thee wee only Red and Geen fish in the pond and Red fish is the fist species to ecome extinct. Thus P (R B +. 5
6 Similaly, P (R G +. P (R P (R BP (B + P (R GP (G Polem 3.T.3 Fist let s pove that C(n, m y n ( y m n!m! dy y induction on m. (n + m +! If m then C(n, y n dy yn+ n + n + n!!. Suppose it is tue fo m. (n + +! Then C(n, m + y n ( y m+ dy y n ( y m dy C(n, m C(n +, m n!(m +! (n + m +! ( y n ( y m y n+ ( y m dy y n+ ( y m dy n!m! (n +!m! (n + m +! (n + m + 2! Now let H and C i e as in Example 5e. Let F n e the event that the fist n flips esulted in heads and n tails. Then P (H F n P (H F n C i P (C i F n i P (H F n C i i i i ( ( n i ( n j ( i i j ( i i P (F n C i P (C i (F n C j P (C j j ( i ( i ( i + ( i n ( i n P (H F n C i P (F n C i P (C i P (F n n + n + C( +, n C(, n + n + 2 6
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