# Geometric Probability > Geometric probability for JEE Mathematics: favourable measure over total measure, meeting problems, Buffon's needle and Bertrand's paradox caution. - Canonical URL: https://prepelephant.com/topics/jee/mathematics/geometric-probability - Exam / course: JEE · Subject: Mathematics - Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team - First published: 2026-10-02 - Last updated: 2026-10-02 - How to cite: "Geometric Probability", PrepElephant, https://prepelephant.com/topics/jee/mathematics/geometric-probability ## Direct answer Probability becomes geometry when outcomes form a continuum: P(event) = favourable measure / total measure, where the measure is length, area or volume depending on the dimension of the sample space. A chord chosen by a point in a circle, a meeting time chosen in an hour, a point thrown into a square — each replaces counting with measurement. The archetype is the waiting-friends problem: two friends arrive independently and uniformly within an hour and each waits ten minutes; the probability they meet is 1 − (50/60)² = 11/36, the complement of the unshaded triangle in the unit square of arrival times. The method's health depends entirely on the uniformity of the measure — Bertrand's paradox shows that "random chord" is meaningless until the randomisation procedure is pinned down. ## What you must remember - **Core formula:** P = favourable length (or area, volume) / total length (area, volume), valid when the distribution is uniform over the region. - **Meeting problem:** arrivals uniform on [0, T], waiting time t: P(meet) = 1 − (1 − t/T)²; with T = 60 and t = 10 minutes, 11/36. - **Buffon's needle:** a needle of length l dropped on lines spaced d (l ≤ d) crosses a line with probability 2l/(πd) — historically the experimental route to estimating π. - **Points in regions:** P(point lands in sub-region) = area ratio; a point uniform in a square of side 2 lying inside the inscribed circle has probability π/4 — the Monte Carlo estimator of π. - **Two-variable strategy:** draw the sample space in the (x, y) plane, mark the event as a region defined by the condition (often |x − y| < t), compute the area ratio. - **One-dimension version:** intervals on a segment: probability a randomly chosen point in [0, a] lies within [b, c] is (c − b)/a. - **Bertrand's warning:** "a random chord of a circle" yields 1/3, 1/2 or 1/4 depending on the construction — JEE questions always specify the mechanism (random midpoint, random endpoints, random distance from centre), and so should any solution. ## The waiting-friends problem Two friends agree to meet at a cafe between 6 and 7 pm; each arrives at a uniformly random instant and leaves after 10 minutes. Let x and y be arrival minutes after 6. The sample space is the 60 × 60 square. They meet exactly when |x − y| ≤ 10, a diagonal band of width 10√2 around the line y = x. The complement — the two triangles where |x − y| > 10 — has combined area (50 × 50)/2 × 2 = 1250. So P(meet) = 1 − 1250/3600 = 2350/3600 = 11/36. Every parameter change is now free: waiting 15 minutes gives 1 − (45/60)² = 7/16; waiting the full hour guarantees the meeting. The drawing step is the entire method. Conditions translate to regions: |x − y| < t is a band; x + y < T is a half-plane cut by a diagonal; x² + y² < r² is a disc. Once the region is shaded, the probability is bookkeeping — and the examiner's real question was whether you could translate English into a region. ## Assumptions that decide the answer The universal trap is non-uniformity: if arrival times cluster near the start of the hour, the square is no longer homogeneous and area ratios lie. JEE questions state "uniformly" precisely because of this. The second trap is measure mismatch — computing a length ratio when the experiment throws a point into an area, or forgetting that a "random point in a circle" carries radial density proportional to r when using polar coordinates (uniform in the disc is not uniform in (r, θ)). Main-level problems are single-region ratios; Advanced composes conditions — meeting in a particular order, or the waiting time that hits a specified probability. Bertrand's paradox itself is quotable in interviews: it is the standard demonstration that geometric probability begins in the modelling, not the arithmetic. ## Frequently asked questions ### What is the formula for geometric probability? P(event) = favourable measure / total measure, with measure being length, area or volume, assuming outcomes are uniformly distributed. ### What is the probability that two friends waiting 10 minutes each meet within a 60-minute window? 11/36: the complement region |x − y| > 10 consists of two triangles of total area 1250 out of 3600. ### What does Buffon's needle experiment measure? The probability that a needle of length l (l ≤ d) crosses one of parallel lines spaced d apart is 2l/(πd), so repeated throws estimate π. ### Why is "a random chord" ambiguous? Bertrand's paradox: chords defined by random endpoints, random midpoints or random distance from centre give 1/3, 1/2 and 1/4 respectively, so the selection procedure must be specified. ### Is a point uniform in a circle uniform in polar coordinates? No — uniform in the disc means the radial coordinate has density proportional to r; treating r as uniform overstates the outer region.