 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-2/16/a/i/solution

Let

X

0, 1, 2

, and counting equally likely three-element samples gives

P (X = k) = \frac{( k 2 ) ( 3 - k 3 )}{( 3 5 )} .

Therefore

P (X = 0) = \frac{1}{10}, P (X = 1) = \frac{3}{5}, P (X = 2) = \frac{3}{10} .

Their sum is

(1 + 6 + 3) /10 = 1

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