 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ib/paper-2/6h/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Under fairness every expected count is

90/6 = 15

. Therefore

X^{2} = \frac{( 20 - 15 ) ^{2} + ( 15 - 15 ) ^{2} + ( 12 - 15 ) ^{2} + ( 17 - 15 ) ^{2} + ( 9 - 15 ) ^{2} + ( 17 - 15 ) ^{2}}{15} = \frac{25 + 0 + 9 + 4 + 36 + 4}{15} = 5.2 .

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