A dart is randomly thrown at a circular board on which two concentric rings of radii $R$ and $2R$ having the same width (width much less than $R$) are marked. The probability of the dart hitting the smaller ring is:
$(1)$ Twice the probability that it hits the larger ring.
$(2)$ Half of the probability that it hits the larger ring.
$(3)$ Four times the probability that it hits the larger ring.
$(4)$ One-fourth the probability that it hits the larger ring.
My attempt $:$ Let $S$ be the radius of the circular board. Then the probability
that the dirt will hit the larger ring and the smaller ring would be respectively $\frac {4R^2} {S^2}$ and $\frac {R^2} {S^2}.$ So the probability of the dart hitting the smaller square will be one-fourth the the probability that it hits the larger square. So according to me option $(4)$ is correct. But the answer key suggests that the correct option is $(2).$ Am I missing something? Any help in this regard will be appreciated.
Thanks for your time.
Source $:$ CSIR NTA NET DECEMBER $2019.$
Best Answer
Let's say the common width was $W$. Then the area occupied by a ring of radius $r$ would be $$\pi (r+W)^2-\pi r^2=2\pi r W+\pi W^2$$
Thus the probability that the dart will land in that ring is $$\frac {2\pi r W+\pi W^2}{\pi S^2}=\frac {2rW}{S^2}+\frac {W^2}{S^2}\approx \frac {2rW}{S^2}$$ where we have used the fact that $W$ is much smaller than $S$.
It follows that the ratio you want is approximately $$\frac {2RW}{S^2}\Bigg /\frac {4RW}{S^2}=\frac 12$$
As desired.