IMO A...but if have to solve this within 2min, I'd choose E as soon as I read the question...nasty question.. took me over 5min to think what I should do.
Let single = s, homerun=r, strikeout=o
S1. Gs = 2Rs and Go = 4Ro
possible scenarios not to score a point before strikeout
!P(G) = Go+Gs*Go+Gs*Gs*Go
!P(R) = Ro+Rs*Ro+Rs*Rs*Ro = Ro(1+Rs+Rs^2)
!P(G) = 4Ro+2Rs*4Ro+2Rs*2Rs*4Ro = 4Ro(1+2Rs+4Rs^2)
!P(G)-!P(R) = 4Ro(1+2Rs+4Rs^2) - Ro(1+Rs+Rs^2)
= Ro(3+7Rs+15Rs) > 0
Thus, Roger is more likely to allow a score. sufficient.
S2. Gs=2Rs and Gh=(1/4)Rh
possible scenarios to score a point before strikeout
P(G) = Gs*Gs*Gs*Go + Gs*Gs*Gh*Go + Gs*Gh*Go + Gh*Go
= 2Rs*2Rs*2Rs*Go + 2Rs*2Rs*(1/4)Rh*Go + 2Rs*(1/4)Rh*Go + (1/4)Rh*Go
P(R) = Rs*Rs*Rs*Ro + Rs*Rs*Rh*Ro + Rs*Rh*Ro + Rh*Ro
Seems we need Ro and Go to solve this.. insufficient not 100% sure
The only thing that matters is what you believe.