 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1/16/a/v/solution

Let

I_{j}

be the indicator random variable of a black ball at draw

j

. The black count after three draws is

N (B_{3}) = I_{1} + I_{2} + I_{3}

E [N (B_{3})] = j = 1 \sum 3 E [I_{j}] = 3 \frac{4}{7} = \frac{12}{7} .

Independence is not required for this step. This is also the mean of the corresponding hypergeometric distribution.

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