Question

Two points on the circumference of a circle of radius $1$ are chosen at random. Find the probability that the distance between the two points is at most $1 .$
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VtProblem 4 (47691)
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The numbers $x_1,$ $x_2,$ $x_3,$ $x_4$ are chosen at random in the interval $[0,1].$ Let $I$ be the interval between $x_1$ and $x_2,$ and let $J$ be the interval between $x_3$ and $x_4.$ Find the probability that intervals $I$ and $J$ overlap.
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VtProblem 5 (49182)
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Let $O$ be the center of square $ABCD.$ A point $P$ is chosen at random inside the square. Find the probability that $P$ is closer to $O$ than to any of the vertices. (In other words, find the probability that $OP$ is shorter than $AP,$ $BP,$ $CP,$ and $DP.$)
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VtProblem 6 (48985)
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Right triangle $XYZ$ has legs of length $XY = 12$ and $YZ = 10$. Point $D$ is chosen at random within the triangle $XYZ$. What is the probability that the area of triangle $XYD$ is at most $24$?
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VtProblem 7 (49808)
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The grid below is made up of line segments, like the line segment in red.

[asy]
unitsize(1 cm);

int i;

for (i = 0; i <= 4; ++i) {
draw((i,0)--(i,2));
}

for (i = 0; i <= 2; ++i) {
draw((0,i)--(4,i));
}

draw((4,1)--(4,2), linewidth(2*bp) + red);

label("$A$", (0,0), SW);
label("$B$", (4,2), NE);
[/asy]

There are a number of paths that go from $A$ to $B$ in the grid, where every step goes to the right or up. If we choose a line segment at random, then what is the expected number of paths that pass through it?
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VtProblem 8 (44682)
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A stick is broken at two points, chosen at random. If the length of the stick is $6$ units, then what is the probability that all three resulting pieces are shorter than $5$ units?
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Answer 1

Here are brief solutions for each problem:

1) Two random points on unit circle — P(distance ≤ 1)
- Let φ be the (minor) central angle between the points; φ is uniform on [0, π].
- Chord length = 2 sin(φ/2) ≤ 1 ⇒ sin(φ/2) ≤ 1/2 ⇒ φ/2 ≤ π/6 ⇒ φ ≤ π/3.
- Probability = (π/3)/π = 1/3.

2) Four random numbers x1,x2,x3,x4 ∈ [0,1]; I = interval between x1,x2, J between x3,x4. P(I and J overlap)
- Order the four distinct points: there are 4! = 24 equally likely orderings. I and J do not overlap exactly when both endpoints of one interval are the two leftmost or the two rightmost points. That gives 8 orderings.
- So overlap probability = 1 − 8/24 = 16/24 = 2/3.

3) Point P uniformly in square ABCD with center O. P(OP < distance to any vertex)
- The locus where OP < distance to each vertex is the intersection of four half-planes, giving the diamond (rotated square) whose vertices are the midpoints of the sides of ABCD. Its area is half the area of ABCD.
- Probability = 1/2.

4) Right triangle XYZ with legs XY = 12 (base) and YZ = 10 (height). D random in triangle. P(area(XYD) ≤ 24)
- Area of XYD = (1/2)*12*h = 6h where h is distance from D to base XY. So 6h ≤ 24 ⇒ h ≤ 4.
- The proportion of triangle within distance 4 of the base = 1 − (remaining top small triangle area / total) = 1 − (6/10)^2 = 1 − 36/100 = 64/100 = 16/25.

5) Grid from A(0,0) to B(4,2). Choose a segment uniformly at random; expected number of monotone (right/up) A→B paths that pass through it.
- Total monotone paths = C(6,2) = 15. Each path uses 6 segments, so total path-segment incidences = 15·6 = 90.
- Number of segments in the grid: vertical 5 columns × 2 segments = 10, horizontal 3 rows × 4 segments = 12, total = 22.
- Expected number = 90/22 = 45/11.

6) Stick length 6 broken at two random points. P(all three pieces < 5)
- In the (ordered) break-point triangle 0 ≤ X ≤ Y ≤ 6 (area 18), the allowed region is {Y > 1, X < 5, Y − X < 5}. Integrating gives allowed area = 16.5.
- Probability = 16.5/18 = 11/12.

Summary of final answers:
1) 1/3
2) 2/3
3) 1/2
4) 16/25
5) 45/11
6) 11/12