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 concise solutions to each problem.

Problem 1 (circle, chord ≤ 1)
- Let θ be the smaller central angle between the two random points. θ is uniform on [0, π]. The chord length is 2 sin(θ/2).
- 2 sin(θ/2) ≤ 1 ⇔ sin(θ/2) ≤ 1/2 ⇔ θ/2 ≤ π/6 ⇔ θ ≤ π/3.
- Probability = (π/3)/π = 1/3.

Problem 2 (two random intervals in [0,1])
- The intervals fail to overlap exactly when {x1,x2} are both to the left of {x3,x4} or vice versa. By symmetry, probability of one specified ordering (x1,x2 both smallest) = (2!·2!)/4! = 1/6. Doubling gives no-overlap = 1/3.
- Overlap probability = 1 − 1/3 = 2/3.

Problem 3 (point in square closer to center than to any vertex)
- Place square vertices at (±1,±1), center at (0,0). The perpendicular bisectors between O and the four vertices are the lines x±y = ±1. Their intersection is the diamond with vertices (±1,0),(0,±1).
- Area of the diamond = (diagonal1·diagonal2)/2 = (2·2)/2 = 2. Square area = 4.
- Probability = 2/4 = 1/2.

Problem 4 (random point in right triangle, area(XYD) ≤ 24)
- Put Y at (0,0), X at (12,0), Z at (0,10). For D = (x,y) in triangle, area(XYD) = (1/2)·12·y = 6y. So area ≤ 24 ⇔ y ≤ 4.
- Area of region 0 ≤ y ≤ 4 inside triangle = ∫0^4 (12 − (6/5)y) dy = 192/5. Total triangle area = 60.
- Probability = (192/5)/60 = 16/25.

Problem 5 (expected number of A→B monotone paths through a random segment)
- Number of monotone paths from A to B (4 rights, 2 ups) = C(6,2) = 15. Each path uses exactly 6 segments, so total path-segment incidences = 15·6 = 90.
- Total number of unit segments in the grid = vertical: 5·2 = 10, horizontal: 3·4 = 12, total 22.
- Expected number = 90/22 = 45/11.

Problem 6 (stick length 6 broken at two random points; all pieces < 5)
- Let breakpoints be uniform on [0,6]^2. The event a given piece ≥5 has area 1 in the 36-area square for each piece; these events are disjoint. So probability some piece ≥5 = 3·(1/36) = 1/12.
- Desired probability = 1 − 1/12 = 11/12.

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