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A point is chosen at random inside a circle. What is the probability that the point is closer to the centre of the circle than to its boundary?
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Verified Answer
The correct answer is:
$\frac{1}{4}$
Probability $=\frac{\pi\left(\frac{\mathrm{r}}{2}\right)^{2}}{\pi \mathrm{r}^{2}}$
$=\frac{\pi\left(\frac{r^{2}}{4}\right)}{A r^{2}}$
$=\frac{r^{2}}{4 r^{2}}=\frac{1}{4}$
$=\frac{\pi\left(\frac{r^{2}}{4}\right)}{A r^{2}}$
$=\frac{r^{2}}{4 r^{2}}=\frac{1}{4}$
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