Bob is a recent law Sch ool Fr ad uates who intends tO take the state bare ex am According to the National conferenee Examiners, ab out 45 % of all PeoPle who take the past, Let state ba exam re present the number of times n=1,2,3, .-. t okes he bar exam untill Per som the first pass vari abl e X pandom de fine Le t re Pre sents the number of times the bar exam Un till Person takes the first pass the de fine P re presents take! the State Probability of a Person bar exam. Jeometrie distribuhion follow So, with Parame ter P f ake who 457. of all people Here bar exam. state P = 0.45 the So, X ~ Geometrie (P) So, X v Geometrie (o.45) Pmf of x P(x-n) = - P(I-P) The n- P(X= m) = 6. 45) · Co.55) Or, ; n=1,2,3,-..
or, P(m) (6-45) (0.55) n= 1, 2, 3, - - - . (b) what is the Probability that Bob first the bap exam passes the second try SOIUHon : Bob first pa sses the bar exam on the second try.(n=2) PC x= 2) (- 45) (0.55)* 2-) j.e, = 0.2475 -0:248 that Bob is the what Pro bability attempts three to pass the needs exam1. So lution: firs t passes the bar ex am om Bob (n= 3) try third 3-) P(x=3) = 6:45) (0.55) =0.1361 25 *0.1 36
what is the Probability that Bob needs more than attempts to pass three the exami .. Solution: number of try more than three. Here m 73) í.e, P( x > 3) |- P( x <3) %3D PCX=2)- PCx=3) (- P(X =1) - = |- (- 45)-(0.55) - (0.45) (0.55)* - (0. 45)-(0.55) 1-1 2-1 3-1 1- (0.45) - k. 45 x 0 - 55)- (0.45 xo.557) O. 166 375 expec ted number what is Hhe Bob State.barrexam attempts at the fom his. firse) Pass must make So1utiom of geometrie ex peeted value The is, dis tribution n-1 I n.p(-P) E(x)
Pit 2P(rP) + 3 P(HP) E6) = multiPly e4© by (1-P), wegel (1- P) ELX) = PLI-P) + 2pC-P) sub strueting eq© from eg@, weget, E(x) - (-P) E(X) P+ PI-P) + P(I-P) E(x) - E(x) +PEX) = P[ It (-P) + EP)te on, P E(X) = P %31 or, 1-(1-P) ECX) or, Here, P=0. 45 So, = 2-2222 E(x) 0.45 - 2.222 :The expee ted Nalue is 2- 222