Question 1089244
Among the numbers from 1 to 21,
there are {{{10}}} even numbers, from {{{1*2=2}}} to {{{10*2=20}}}  ;
there are {{{7}}} multiples of 3, from {{{1*3=3}}} to {{{7*3=21}}} ,
including {{{3}}} multiples of 6 (even and multiples of 3), from {{{1*6=6}}} to {{{3*6=18}}} .
Those multiples of 6 would be counted doubly, so among the numbers from 1 to 21,
there are {{{10+7-3=14}}} that are either even and/or multiple of 3.
The probability that the spinner will stop on one of those {{{14}}} numbers is
{{{14/21=highlight(2/3)}}} .