Without loss of generality we could assume that one die is #color(red)("red")# and the other is #color(green)("green")#

For each of the 6 #color(red)("red")# outcomes there are 6 #color(green)("green")# outcomes.

This means that there are a combined collection of #color(red)6xxcolor(green)6=color(blue)(36)# possible outcomes.

Of the #color(blue)(36)# possible outcomes only #color(magenta)3# result in totals greater than on equal to 11; namely

#color(white)("XXX")< color(red)("red"),color(green)("green") > in { < color(red)5, color(green)6 >, < color(red)6, color(green)5 >, < color(red)6, color(green)6 >}#

Therefore, there are #color(blue)(36)-color(magenta)3=color(brown)33# possible outcomes for which the sum is less than 11.

Since each outcome is (assumed to be) equally probable,

#color(white)("XXX")P("sum" < 11) = (color(brown)(33))/(color(blue)(36))=11/12#