Question 473012
No the area of largest circle - area of smallest circle does NOT equal area of 2nd circle.
The largest circle represents the entire target, now if you take away the smallest circle you are left with the entire outer part of the target which includes part of the outer circle as well as part of the 2nd circle.
You are correct in thinking it is 4/9.
P(hitting 2nd circle) = Area of 2nd / Area of target
= {{{(4*pi)/(9*pi) = 4/9}}}
**Note, I am assuming it includes the possibility of hitting the inner circle as well, since it is inside the 2nd circle**
Otherwise, if you only wanted to know the probability of hitting the inner ring (2nd circle but not the smallest circle) then subtract the smallest circle from 2nd circle
P = {{{(4*pi - pi)/(9*pi) = 3*pi/(9*pi) = 3/9 = 1/3}}}