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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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All Answers 1
Answered by
GPT-5 mini
AI
Here are the answers with brief justifications.
Problem 1. Two random points on a circle of radius 1; probability their distance ≤ 1.
- Let Δ be the smaller central angle between the points (0 ≤ Δ ≤ π). The chord length is 2 sin(Δ/2). So 2 sin(Δ/2) ≤ 1 ⇔ sin(Δ/2) ≤ 1/2 ⇔ Δ/2 ≤ π/6 ⇔ Δ ≤ π/3.
- Δ is uniformly distributed over [0,π], so the probability = (π/3)/π = 1/3.
Answer: 1/3.
Problem 2. Four independent uniform points x1,…,x4 in [0,1]; I between x1,x2, J between x3,x4. Probability I and J overlap.
- The intervals are disjoint only if both endpoints of one interval lie to the left of both endpoints of the other. Among the 4! = 24 orderings of the four labeled points, there are 4 orderings with x1,x2 the two smallest (in either order) and x3,x4 the two largest (in either order), so probability = 4/24 = 1/6 that I is entirely left of J. Similarly 1/6 for J entirely left of I. Thus disjoint probability = 1/3.
- Therefore overlap probability = 1 − 1/3 = 2/3.
Answer: 2/3.
Problem 3. Point P uniformly in square ABCD with center O; probability OP < distance to each vertex.
- Place square centered at origin with side 2, vertices (±1,±1). OP^2 ≤ distance to (1,1)² gives x+y ≤ 1; similar inequalities for other vertices give
x+y ≥ −1, x−y ≤ 1, x−y ≥ −1.
- These four inequalities define the diamond with vertices (±1,0),(0,±1), whose area is 2. The square’s area is 4.
- Probability = 2/4 = 1/2.
Answer: 1/2.
Problem 1. Two random points on a circle of radius 1; probability their distance ≤ 1.
- Let Δ be the smaller central angle between the points (0 ≤ Δ ≤ π). The chord length is 2 sin(Δ/2). So 2 sin(Δ/2) ≤ 1 ⇔ sin(Δ/2) ≤ 1/2 ⇔ Δ/2 ≤ π/6 ⇔ Δ ≤ π/3.
- Δ is uniformly distributed over [0,π], so the probability = (π/3)/π = 1/3.
Answer: 1/3.
Problem 2. Four independent uniform points x1,…,x4 in [0,1]; I between x1,x2, J between x3,x4. Probability I and J overlap.
- The intervals are disjoint only if both endpoints of one interval lie to the left of both endpoints of the other. Among the 4! = 24 orderings of the four labeled points, there are 4 orderings with x1,x2 the two smallest (in either order) and x3,x4 the two largest (in either order), so probability = 4/24 = 1/6 that I is entirely left of J. Similarly 1/6 for J entirely left of I. Thus disjoint probability = 1/3.
- Therefore overlap probability = 1 − 1/3 = 2/3.
Answer: 2/3.
Problem 3. Point P uniformly in square ABCD with center O; probability OP < distance to each vertex.
- Place square centered at origin with side 2, vertices (±1,±1). OP^2 ≤ distance to (1,1)² gives x+y ≤ 1; similar inequalities for other vertices give
x+y ≥ −1, x−y ≤ 1, x−y ≥ −1.
- These four inequalities define the diamond with vertices (±1,0),(0,±1), whose area is 2. The square’s area is 4.
- Probability = 2/4 = 1/2.
Answer: 1/2.
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