Question 1183619
1.  P(at least one big prize) = 1 - P(no big prize).

Now P(no big prize) = {{{((matrix(2,1,4,0))*(matrix(2,1,20,20)))/((matrix(2,1,24,20))) = 1/10626}}}


Hence   P(at least one big prize) = {{{1 - 1/10626 = highlight(0.999906)}}} to 6 d.p., which is virtually 1.
In other words, with 20 attempts you almost surely will have at least one big win, or one big prize, however you want to call it.


2. P(all big prizes) = {{{((matrix(2,1,4,4))*(matrix(2,1,20,16)))/((matrix(2,1,24,20))) = highlight(1615/3542)}}} ~ 0.455957 to 6 d.p.