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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. – minopret Dec 21 '12 at 5:28
up vote 5 down vote accepted

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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I was trying to help without being too explicit. Unfortunately, this complicates things sometimes... – copper.hat Dec 21 '12 at 5:56
Very nice straddling delicate line between saying too little and too much. – André Nicolas Dec 21 '12 at 7:11
For part (b), isn't the area described the difference between the area of the square and the two triangles you describe? – idealistikz Dec 21 '12 at 9:31
Good catch, I have fixed the bug... – copper.hat Dec 21 '12 at 16:01
+1 especially for the first sentence. – Dilip Sarwate Dec 21 '12 at 16:49

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