Question

Next weekend Team A plays a 5 game series against their nemesis Team D. Team A has a 65% chance of winning each game, independent of others. Let X denote the number of games that Team A wins in a 5 game series. 6 Use the PDF of X to find the expectation 4 = EX and variance Var(X). 7 Then use a formula from Ch 4.6 to quickly find these. 8 On average, how many games need to be played before Team A wins their first game? How many games until they lose their first game? (See Ch 4.8) 9 Suppose instead that the series continues until Team A wins 3 games. On average, how long will this take? What if it continues until Team D wins 3 games? (See Ch 4.8) 10 Let Y be the event that the series ends in k games. Find the PDF of Y. Then use this to find the expected number of games it takes for the series to finish. Hint: Use the notation WLWW to denote the event that Team A wins game 1, loses game 2, wins 3 and wins 4. This is one way for the series to end in 4 games. Then Y(4) = {LWWW, WLWW, WWLW, WLLL, LWLL, LLWL}

Answer 1

• 6) Ed X z no. of games that team A wins in a games series then x follows a binomial distribution, beat with n=5 and p = 0.65, this Pdf of x is given by: baina (5) 015" 0.335-1 2=0,1,2,3,4,5 hence, E(x) = 3 (3) 0.65* 0.35" = », 045*0.35"> 540.63ž, tas, o sos** det t=n- 4-t asossz, den 0 49 jan - 5*0**325 - Smart Tiyas Binomial (4,0.65) now, ECxor) = -Ossxy F one day, 4*03" } 2016035* 6 065 035 heu= x-2 22 U (3-4)! นะ) = 0.6s'xsxux Š cu . 2.65*0.324 les binomial (8,0.65) rus thus, var (x) = E(X)-ECx)) = E(X(x-11)+F(x)-TEC®,)? +3.25 -(3-2552 1,1375 5X0-65 X 0-3 = ие

8. 0.65 - 1.538 as expectation of Let w denotes the no. of games required to A for winning their first game then w follows geometric (0.65) distribution, hence E(W) = 263 1.538 geometric (p) and if L denotes the no. of games required to A for losing their first game then I follows a geometric (0.35) distribution, hence E(L) = 0.35 = 2.857 Let Gowa denotes the no. of games required for team A wins 3 games. Then WA follows a Negative Binomial (3,0.65) distribution. We know, expectation of negative binomial = 1 Hohence, E(wa) 2os = 4.615 Similarly if Wo denotes no. of games required till Then Wo follows team D wins 3 games. till a negative binomial (3,0.35) distribution Thus E (Wy) = 3 5 - 8.511

Here I denotes the series ends in a gams. Lt. in k games either of Team A and Team D wins 3 games. This values of k={3,4,53 us at least 3 games are required to win 3 games and at almost 5 games are required to win 3 games first. Hence. PCY=3) = P { (www), (1, 1,-) } 0.653 + 0-35° = 03175 Ply=4)= p { Lwww, wzwW, WWLW, WLLL, LWLL, EL W23 3x 0.65 0.35 + 3x 0-65X0-353 = 0.37196 and Ply35)-p{LL www, LWLWW, LWWLW, WLLWW, WLWLW, WWLLW, WWWLLL, WLWLL, WLLWL, LWLWL, LLWWL, LWWLL ģ - = 680.65% 0.35²+ 6x0.65x0.353 0.31054