Question 1043583
The act of playing the game multiple times is a binomial experiment with p = 0.20.  
The probability of winning at least one prize is given by the expression

{{{sum((matrix(2,1,n,k))*0.8^(n-k)*0.2^k, k = 1,n)}}}.

Hence we want {{{1 - 0.8^n = sum((matrix(2,1,n,k))*0.8^(n-k)*0.2^k, k = 1,n) > 0.9}}} or 

{{{1 - 0.8^n > 0.9}}}.

<===> {{{0.10 > 0.8^n}}}

===> {{{ln(0.10) > ln (0.8^n) = n*ln(0.80)}}}, since ln is an increasing function.

===> {{{ln(0.10)/ln(0.80) = 10.31885116 < n}}}, since ln(0.80) < 0.

===> {{{11 <= n}}}.

Therefore least number of games that must be played to ensure that the probability of winning at least one prize is more than 0.90 is {{{highlight(11)}}}.