 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2020/paper-1/16/a/iii/solution

1/2

P (A_{2} ∣ A_{3}) = \frac{P ( A _{3} ∣ A _{2} ) P ( A _{2} )}{P ( A _{3} )} = \frac{( 4/9 ) ( 1/2 )}{1/2} = \frac{4}{9} .

This also agrees with exchangeability of the second and third positions.

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