 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2/14/b/solution

Apply the same dimensionless transformation independently to the two cups, giving

X, Y \in [- π /2, π /2]

. The total-volume condition becomes

X + Y \geq π /2

. Since neither variable can exceed

π /2

, this region has

0 \leq X \leq π /2

and

π /2 - X \leq Y \leq π /2

, the triangle shown above.

\frac{1}{4} cos x cos y

. The desired double integral is

P = \frac{1}{4} \int_{0}^{π /2} cos x \int_{π /2 - x}^{π /2} cos y d y d x = \frac{1}{4} \int_{0}^{π /2} cos x (1 - cos x) d x = \frac{1}{4} [1 - \int_{0}^{π /2} \frac{1 + cos 2 x}{2} d x] = \frac{1}{4} - \frac{π}{16} .

Thus

P (v_{1} + v_{2} \geq 630 mL) = \frac{4 - π}{16} .

Boundary equality has probability zero for these continuous random variables.

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