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A point $(x, y)$ is to be selected from the square $S$ containing all points $(x, y)$ such that $0 \leq x \leq 1$ and $0 \leq y \leq 1$. Suppose that the probability that the selected point will belong to each specified subset of $S$ is equal to the area of that subset. Find the probability of each of the following subsets: ${}{}$

(a) the subset of points such that $(x - \frac{1}{2})^{2} + (y-\frac{1}{2})^{2} \geq \frac{1}{4}$;
(b) the subset of points such that $\frac{1}{2} < x + y < \frac{3}{2}$,
(c) the subset of points such that $y \leq 1 - x^{2}$;
(d) the subset of points such that $x = y$.

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    The probability is equal to the area, as it says. If the area seems difficult to determine, it may help to draw the picture.2012-12-21

1 Answers 1

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In all cases you should draw a picture.

For (a), the set of points described is the unit square with a disk of radius $\frac{1}{2}$ centered at $(\frac{1}{2}, \frac{1}{2})$ removed. Since the disk is contained in the square, you can figure out the area by subtracting the area of the circle from the area of the square.

For (b), the area described is two similar triangles, one with vertices $(0,0), (\frac{1}{2},0), (0,\frac{1}{2})$, and the other with vertices $(1,1), (\frac{1}{2},1), (1,\frac{1}{2})$. The area is the alea of the square less the sum of the two triangles.

For (c), you will need to break out the integration toolkit. The area is that bounded by the $x$-axis, $y$-axis and the curve $y=f(x)=1-x^2$. The area is $\int_0^1 f(x) dx$, but you need to do the integration.

For (d), you should be able to make a guess at the area of an ideal line. If not, you could figure out the area not on the line, and subtract that from the area of the square. The area not on the line is bounded by the triangles $(0,0), (1,1), (0,1)$ and $(0,0), (1,1), (1,0)$.

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    +1 especially for the first sentence.2012-12-21