10! !. 3. Find the probability that a five-card poker hand (i.e. 5 cards out of a 52-card deck) will be:

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1 MATH 0(001 Fall 2018 Homewok 2 Solutions Please infom you instucto if you find any eos in the solutions 1 Suppose that thee ae duck huntes, each with a pefect shot A flock of ducks fly ove, and each hunte selects one duck at andom and shoots Find the pobability that ducks ae killed Solution: Because the shots ae all andom, we ae in the setting of equally likely events Each hunte chooses one of the ten ducks to shoot at, so thee ae possible ways fo the huntes to shoot Thee ae !/! ways fo none of the two huntes to shoot the same duck The pobability is then!! 2 A confeence oom contains m men and w women These people seat at andom in m + w seats aanged in a ow Find the pobability that all the women will be adjacent Solution: Thee ae (m + w! possible ways to seat the m + w people in a ow To detemine how to seat all the women togethe, we can pick how many men we place to the left of the ow of women Ou options ae any numbe in {0, 1, 2, m}, so thee ae m + 1 possible choices We then have to conside all possible odeings of the men, of which thee ae m! choices, and all possible odeings of the women, of which thee ae w! choices The answe is then (m + 1m!w! (m + w! 3 Find the pobability that a five-cad poke hand (ie cads out of a 2-cad deck will be: (a Fou of a kind, that is fou cads of the same value and one othe cad of a diffeent type (xxxxy shape (b Thee of a kind, that is thee cads of the same value and two othe cads of diffeent values (xxxyz shape (c A staight flush, that is five cads in a ow of the same suit (ace may be high o low (d A flush, that is five cads of the same suit, but not a staight flush (e A staight that is five cads in a ow, but not a staight flush (ace may be high o low Solution: Thee ae ( 2 possible hands we can daw fom a 2 cad poke deck

2 (a To pick fou of a kind, we fist choose the ank of the fou cads fom among the anks in the deck We then choose the emaining cad fom the 48 othe cads in the deck The pobability is then 48 ( 2 (b We fist choose the ank of the thee cads that will be the same Thee ae ways to do so Once we know the ank, thee ae ( choices of which thee of these cads we daw Then we choose the anks of the two emaining cads, which must be diffeent Thee ae ( 12 2 ways to do so Fo each of these two anks, we have 4 options fo which cad to choose Thee ae then ( ( 2 (c We fist pick which suit we ae going to choose, of which thee ae 4 choices Then we pick which cad will be low (ie cad with the smallest ank, of which thee ae choices (the only possibilities ae ace o a numbe in {2,, } The pobability is then 40 ( 2 (d We will fist count all flushes and then subtact off the staight flushes, which we just counted To obtain a flush, we fist pick a suit, of which thee ae 4 choices Then we pick which five anks of that suit we will use, of which thee ae ( choices Subtacting off the numbe of staight flushes, we see that the pobability is 4 ( 40 ( 2 (e We will follow the same outline as in pat (d Fist we count all staights and then we subtact off those which ae staight flushes We pick the low cad as in pat (c, of which thee ae choices Then we pick the 4 suits of these five cads (each cad has one of fou suits Subtacting off the numbe of staights that ae also flushes, we see that the pobability is 4 40 ( 2 4 An expeiment consists of dawing cads fom an odinay 2-cad deck (a If the dawing is made with eplacement, find the pobability that no two cads have the same face value (b If the dawing is made without eplacement, find the pobability that at least 9 cads will have the same suit Solution: We will assume thoughout this poblem that the ode of the cads does not matte

3 (a If the dawing is made with eplacement then thee ae 2 /! possible ealizations The numbe of hands in which no two cads have the same face value if we daw without eplacement and ode does not matte is ( 2 possible ways to daw ten cads with no epetitions The pobability is then 2! 42!2 (b Without eplacement, thee ae ( 2 ways to select cads fom the deck If we want at least nine of the cads to have the same suit we will beak the poblem into picking exactly nine and exactly ten cads of the same suit In eithe case, we fist pick the suit, of which thee ae 4 choices To pick exactly nine cads, we then choose 9 anks fom among the anks and then pick a cad fom the 2 39 cads emaining in the deck Thee ae ( ( 9 39 ways to do this Similaly, thee ae ways to choose ten cads of any given suit The pobability is then 4 ( 9 ( ( 2 An un contains balls numbeed 1 to We daw five balls fom the un, without eplacement Find the pobability that the second lagest numbe dawn is 8 Solution: Ode does not matte fo this poblem, so thee ae ( possible choices of balls To have 8 be the second highest choice, we need to pick one ball fom among {9, } and thee balls fom {1, 2, 3, 4,, 6, 7} Thee ae 2 ways to do the fome and ( 7 3 ways to do the latte The pobability is then 2 (7 3 ( 6 Find the coefficient of x in (2 + 3x 8 Solution: Fom the binomial theoem, the coefficient of x is ( (The game of enconte An un contains n tickets numbeed 1, 2,, n The tickets ae shuffled thooughly and then dawn one by one without eplacement If the ticket numbeed appeas in the -th dawing, this is denoted as a match (Fench: enconte Show that the pobability of at least one match is 1 1 2! + 1 3! + ( 1n 1 1 e 1 as n Solution: Let s let A i be the event that we have at match at the i th dawing We ae inteested in computing the pobability of at least one of the events A i occuing This is P ( n A i By the inclusion-exclusion fomula (see page 18 of the notes, we have P A i ( 1 i 1 P (A j1 A ji 1 j 1 <j 2 < <j i 1 <j i

4 So now we just need to compute what P (A j1 A ji is, fo all choices of subsets j 1 < j 2 < < j i taken fom{1,, n} without eplacement Note that given any subset {k 1, k i } of size i taken fom {1,, n} without eplacement and whee ode does not matte, we can always ode them so that k 1 < k 2 < < k n Fo counting puposes, we see that thee ae the same numbe of subsets {j 1,, j i } of {1, 2,, n} which satisfy j 1 < j 2 < < j i as thee ae subsets taken fom {1, 2,, n} without eplacement and whee ode does not matte That is, thee ae i tems in the second sum above The actual expeiment hee of dawing n numbes and then ecoding those values is the same as dawing n numbes fom the set {1, 2,, n} without eplacement, whee ode mattes Thee ae ways to do this Now suppose that I fix the numbe i The event A j1 A ji means that on the j1 th daw, I pick the cad labelled j 1, on the j2 th daw, I pick the cad labelled j 2, and so on up to j i Fo example, suppose that n, i 3, j 1 1, j 2 3, j 3, then A j1 A j2 A j3 is the event that the fist cad I daw is the cad labelled 1, the thid cad I daw is the cad labelled 3 and the fifth cad I daw is the cad labelled Notice that I have not said anything about what happens to the othe cads, which can be in any configuation (thee can even be moe matches Since we ae fixing i of the values, thee ae (n i! ways to aange the emaining cads Impotantly this pobability is the same fo all choices of j 1,, j i Fom this, we see that fo any i and fo any choice j 1,, j i with j 1 < j 2 < < j i, we have Theefoe, P (A j1 A ji (n i! P A i ( 1 i 1 P (A j1 A ji 1 j 1 <j 2 < <j i 1 <j i ( n (n i! ( 1 i 1 i ( 1 i 1 (n i! i!(n i! ( 1 i 1 1 i! 1 1 2! + 1 3! + + ( 1n 1 8 Show that ( n + m 0 ( m + ( ( n m ( m 0, whee 0 min(n, m and, m, n N algebaic poof Ty to find a combinatoial poof and an Solution: The left hand side is the numbe of ways to fom a commitee of size fom n men and m women The ight hand side is the sum of the sum of the numbe of ways to fom a committee with i men and i women as i anges fom 0 to, so they both count the same thing and theefoe ae equal

5 Algebaically, by the binomial theoem we have We also have (1 + x m+n m+n 0 + m x (1 + x m+n (1 + x m (1 + x n ( m ( ( m x a a a0 b0 m+n 0 j0 ( n x b b ( ( m n x j j Compaing the coefficients of x fo min(m, n, the esult follows