Question 88973
You didn't say how many balls are selected. Lets say you choose 5 balls, and since order doesn't matter, we can use the combination formula:




*[Tex \LARGE \textrm{_{n}C_{r}=\frac{n!}{(n-r)!r!}}] Start with the given formula


*[Tex \LARGE \textrm{_{54}C_{5}=\frac{54!}{(54-5)!5!}}] Plug in {{{n=54}}} and {{{r=5}}}


*[Tex \LARGE \textrm{_{54}C_{5}=\frac{54!}{49!5!}}] Subtract {{{54-5}}} to get 49


*[Tex \LARGE \textrm{_{54}C_{5}]={{{(54*53*52*51*50*49*48*47*46*45*44*43*42*41*40*39*38*37*36*35*34*33*32*31*30*29*28*27*26*25*24*23*22*21*20*19*18*17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2*1)/(49!5!)}}} Expand 54!


*[Tex \LARGE \textrm{_{54}C_{5}]={{{(54*53*52*51*50*49*48*47*46*45*44*43*42*41*40*39*38*37*36*35*34*33*32*31*30*29*28*27*26*25*24*23*22*21*20*19*18*17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2*1)/((49*48*47*46*45*44*43*42*41*40*39*38*37*36*35*34*33*32*31*30*29*28*27*26*25*24*23*22*21*20*19*18*17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2*1)5!)}}} Expand 49!


*[Tex \LARGE \textrm{_{54}C_{5}]={{{(54*53*52*51*50*cross(49*48*47*46*45*44*43*42*41*40*39*38*37*36*35*34*33*32*31*30*29*28*27*26*25*24*23*22*21*20*19*18*17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2*1))/((cross(49*48*47*46*45*44*43*42*41*40*39*38*37*36*35*34*33*32*31*30*29*28*27*26*25*24*23*22*21*20*19*18*17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2*1))5!)}}} Cancel like terms



*[Tex \LARGE \textrm{_{54}C_{5}]={{{(54*53*52*51*50)/5!)}}} Simplify


*[Tex \LARGE \textrm{_{54}C_{5}\frac{379501200}{5!}] Calculate 54*53*52*51*50 to get 379,501,200


*[Tex \LARGE \textrm{_{54}C_{5}=\frac{379501200}{120}] Calculate 5! to get 120



*[Tex \LARGE \textrm{_{54}C_{5}=\frac{379501200}{120}}] Multiply the values in the denominator



*[Tex \LARGE \textrm{_{54}C_{5}=3162510] Divide 


So 54 choose 5 (where order does not matter) yields 3,162,510 unique combinations



So the probability of winning is {{{1/3162510}}}