Single-Variable-Calculus · Unit 19 · Video 3 · Interactive Practice

Probability as a Weighted Average: Part over Whole

IKey Formulas

FormulaMeaningWhat you need
P(x1<x<x2)=∫x1x2w(x) dx∫abw(x) dx\displaystyle P(x_1 < x < x_2) = \frac{\int_{x_1}^{x_2} w(x)\,dx}{\int_a^b w(x)\,dx}Probability = part over wholeA weight w(x)≥0w(x) \ge 0 on the whole range [a,b][a, b]
P ⁣(x>12)=∫1/21(1−x2) dx∫−11(1−x2) dx=5/244/3=532\displaystyle P\!\left(x > \tfrac12\right) = \frac{\int_{1/2}^{1} (1 - x^2)\,dx}{\int_{-1}^{1} (1 - x^2)\,dx} = \frac{5/24}{4/3} = \frac{5}{32}A point chosen by area under y=1−x2y = 1 - x^2The height of the curve is the weight
∫1/211 dx∫−111 dx=14\displaystyle \frac{\int_{1/2}^{1} 1\,dx}{\int_{-1}^{1} 1\,dx} = \frac14The same interval with xx chosen evenlyWeight w(x)=1w(x) = 1

Key Insight: A probability is the weighted average of a function that is 11 on the part you care about and 00 elsewhere. The weight need not total 11: dividing by the whole does the normalizing, and because w≥0w \ge 0 the answer always lies between 00 and 11.

IIVisualization 1 — Random Points under the Arch

Points dropped uniformly into the region under y=1−x2y = 1 - x^2 land to the right of x=cx = c with probability part over whole.

IIIVisualization 2 — Same Interval, Different Weight

The interval [x1,x2][x_1, x_2] keeps its share of the width, but its probability depends on the weight above it.

IVQuiz Questions

Problem 1 · Between the Centre and One Half

Given: a point is chosen at random, by area, in the region under y=1−x2y = 1 - x^2 above the xx-axis — find P ⁣(0<x<12)P\!\left(0 < x < \tfrac12\right).

✅ Correct! Part ∫01/2(1−x2) dx=1124\int_0^{1/2}(1 - x^2)\,dx = \tfrac{11}{24}, whole 43\tfrac43: 1124⋅34=1132\tfrac{11}{24} \cdot \tfrac34 = \tfrac{11}{32}, more than 14\tfrac14 because the curve is tall near the centre.
❌ That picks xx evenly. The interval is a quarter of the width, but choosing by area weights each xx by the height 1−x21 - x^2.
❌ That is the part alone. 1124\tfrac{11}{24} is the area of the slice; divide by the whole area 43\tfrac43.
❌ That is the slice from 12\tfrac12 to 11. This question asks for 0<x<120 < x < \tfrac12, the taller slice next to the centre.
❌ Not quite. Integrate 1−x21 - x^2 from 00 to 12\tfrac12 and divide by the integral from −1-1 to 11.
Show solution
∫01/2(1−x2) dx=[x−x33]01/2=12−124=1124,∫−11(1−x2) dx=43.\int_0^{1/2} (1 - x^2)\,dx = \left[x - \frac{x^3}{3}\right]_0^{1/2} = \frac12 - \frac1{24} = \frac{11}{24}, \qquad \int_{-1}^{1}(1 - x^2)\,dx = \frac43. P ⁣(0<x<12)=11/244/3=1124⋅34=1132.P\!\left(0 < x < \tfrac12\right) = \frac{11/24}{4/3} = \frac{11}{24}\cdot\frac34 = \frac{11}{32}.

Together with P ⁣(x>12)=532P\!\left(x > \tfrac12\right) = \tfrac{5}{32} this gives 1632=12=P(x>0)\tfrac{16}{32} = \tfrac12 = P(x > 0), as symmetry requires.

Problem 2 · A Triangle-Shaped Weight

Given: xx is chosen in [0,2][0, 2] with weight w(x)=xw(x) = x — find P(x<1)P(x < 1).

✅ Correct! ∫01x dx∫02x dx=1/22=14\dfrac{\int_0^1 x\,dx}{\int_0^2 x\,dx} = \dfrac{1/2}{2} = \dfrac14: the left half of the interval carries little weight.
❌ That is the even-weight answer. Half the width, but the weight xx is small there.
❌ That is P(x>1)P(x > 1). The heavy part is to the right of 11; the question asks for the left part.
❌ A probability cannot exceed 11 here. ∫01x dx=12\int_0^1 x\,dx = \tfrac12 is the part; it must be divided by the whole, ∫02x dx=2\int_0^2 x\,dx = 2.
❌ Not quite. Part over whole with the weight xx.
Show solution
P(x<1)=∫01x dx∫02x dx=1/22=14.P(x < 1) = \frac{\int_0^1 x\,dx}{\int_0^2 x\,dx} = \frac{1/2}{2} = \frac14.

Geometrically: the region under w=xw = x on [0,2][0, 2] is a triangle, and the small triangle over [0,1][0, 1] has a quarter of its area.

Problem 3 · Under an Upward Parabola

Given: a point is chosen at random, by area, in the region under y=x2y = x^2 for 0≤x≤10 \le x \le 1 — find P ⁣(x>12)P\!\left(x > \tfrac12\right).

✅ Correct! Part 13−124=724\tfrac13 - \tfrac1{24} = \tfrac{7}{24}, whole 13\tfrac13: 724⋅3=78\tfrac{7}{24} \cdot 3 = \tfrac78.
❌ That picks xx evenly. Chosen by area, the weight is the height x2x^2, which is largest near x=1x = 1.
❌ That is the part alone. Divide the area 724\tfrac{7}{24} by the whole area 13\tfrac13.
❌ That is P ⁣(x<12)P\!\left(x < \tfrac12\right). The left slice has area 124\tfrac1{24}, an eighth of the whole.
❌ Not quite. Integrate x2x^2 over [12,1][\tfrac12, 1] and over [0,1][0, 1], then divide.
Show solution
∫1/21x2 dx=[x33]1/21=13−124=724,∫01x2 dx=13.\int_{1/2}^{1} x^2\,dx = \left[\frac{x^3}{3}\right]_{1/2}^{1} = \frac13 - \frac1{24} = \frac{7}{24}, \qquad \int_0^1 x^2\,dx = \frac13. P ⁣(x>12)=7/241/3=78.P\!\left(x > \tfrac12\right) = \frac{7/24}{1/3} = \frac78.

Problem 4 · Rescaling the Weight

Given: the weight 1−x21 - x^2 on [−1,1][-1, 1] is replaced by 34(1−x2)\tfrac34(1 - x^2), which has total 11 — what happens to P ⁣(x>12)P\!\left(x > \tfrac12\right)?

✅ Correct! The factor 34\tfrac34 multiplies both the part and the whole, so it cancels: only the shape of the weight matters.
❌ The factor appears twice. It scales the numerator and the denominator alike, and cancels in the ratio.
❌ 524\tfrac{5}{24} is the part under the old weight. It is not a probability until divided by the whole 43\tfrac43; under the new weight the part is 34⋅524=532\tfrac34 \cdot \tfrac{5}{24} = \tfrac{5}{32} and the whole is 11.
❌ 14\tfrac14 is the even-weight answer. Rescaling changes the size of the weight, not its shape, so the middle still counts more than the ends.
❌ Not quite. Write out part over whole with the new weight and look for a common factor.
Show solution
∫1/2134(1−x2) dx∫−1134(1−x2) dx=34⋅52434⋅43=5/244/3=532.\frac{\int_{1/2}^{1} \tfrac34(1 - x^2)\,dx}{\int_{-1}^{1} \tfrac34(1 - x^2)\,dx} = \frac{\tfrac34 \cdot \tfrac{5}{24}}{\tfrac34 \cdot \tfrac43} = \frac{5/24}{4/3} = \frac{5}{32}.

Any positive multiple of a weight gives the same probabilities. The rescaled weight just makes the denominator equal to 11.

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