When a point lands "at random", probability turns into geometry: favourable area over total area — or length over length on a line.
The problem
A point is chosen at random inside the square. What is the probability it lands inside the circle?
P = circle area ÷ square area = 25π100 = π4 ≈ 0.785 — about 79%, which matches how much of the square the circle visibly fills. B forgets the π; C would exceed 1, which no probability can.
P = favourable regiontotal region
areas in 2D, lengths in 1D
"At random" with infinitely many possible points means counting fails — so measuring takes over. The same fraction logic as coin problems, with area playing the role of "number of outcomes".
circle: π(5)² = 25π · square: 10² = 100
the inscribed circle's radius is half the side
P = 25π100 = π4
≈ 0.785 — the π stays in the answer
x random on [0, 10]: P(x < 3) = 310
favourable length over total length
P(4 < x < 7) = 310
only the WIDTH of the window matters
A target circle with half the radius covers a QUARTER of the area — (12)² — so hitting it is 14 as likely, not 12. This is the k² scaling rule wearing a probability costume: whenever regions are similar, square the length ratio before it becomes odds.
target: π(5²) = 25π
the whole board
ring: 25π − π(2²) = 21π
outer minus inner
P = 21π25π = 2125
the π cancels
The same subtraction handles any shape cut out of another: a circle minus an inscribed square, a square minus a quarter circle.
square in a circle of radius 4: diagonal = 8, area = 8² ÷ 2 = 32
the diagonal is the diameter
P(inside the square) = 32 ÷ 16π = 2π
outside it: 1 − 2π
6 by 12 rectangle minus a quarter circle of radius 6: 72 − 9π
a quarter of π(6²) = 9π
P = (72 − 9π) ÷ 72 = 8 − π8
divide by the whole rectangle
(5 − 2) ÷ 5 = 35 compares widths. Areas grow with the square of the length, so always subtract areas.
the region is a smaller square
a strip of 1 is trimmed off each side
its side: 10 − 2 = 8
trimmed at both ends
P = 8²10² = 64100 = 1625
Turn every condition into a shape before computing. “Closer to the centre of a circle than to its rim” is the same kind of move: its boundary is the circle with half the radius.
10 − 1 = 9 forgets that the strip comes off both ends of each side.
The window runs off the edge. x is chosen at random from [0, 10]. What is the probability that x is within 3 of 2?
within 3 of 2: from −1 to 5
2 − 3 and 2 + 3
inside [0, 10]: from 0 to 5, length 5
−1 to 0 cannot happen
P = 510 = 12
favourable length over total
What is different: the window sticks out past 0, so only the part inside the interval counts.
A window of width 6 gives 610 only when all of it fits inside [0, 10]. Draw the number line and cut the window at the ends.
A line through a rectangle. A point (x, y) is chosen at random in the rectangle 0 ≤ x ≤ 6, 0 ≤ y ≤ 3. What is the probability that y > x?
y = x meets the top edge at (3, 3)
not at the far corner (6, 3)
region: triangle (0, 0), (0, 3), (3, 3), area 12 × 3 × 3 = 92
above the line
P = (92) ÷ 18 = 14
rectangle area 6 × 3 = 18
What is different: in a square, y = x cuts the area in half; in a rectangle it does not. Find where the line meets the edges before naming the shape.
x + y ≤ 3 lies below the line through (3, 0) and (0, 3)
a right triangle at the origin
triangle: 12 × 3 × 3 = 92
legs of 3
square: 4 × 4 = 16
P = 9216 = 932
3 × 3 = 9 is a square on the legs, not the triangle; 916 is twice the right answer.
Expect a dartboard ring, a shape inside a shape (circle in square, square in circle), a waiting-time or number-line question, or a region in the plane such as x + y ≤ 1. Draw the favourable region, find its area by subtraction if needed, and divide by the whole. Ten-second check: the answer is between 0 and 1, and a length ratio where areas are needed is the usual wrong choice.
The problem
A dartboard is a square of side 12; the bullseye is a centred square of side 4. A dart lands at random on the board. P(bullseye)?
Areas, not sides: 4²12² = 16144 = 19. A is the side ratio 412 — the exact trap from the lesson: similar squares, so the length ratio 13 must be squared before it becomes probability.
