A bag contains 10 white and 3 black balls. Balls are drawn without replacement till all the black balls are drawn What is the probability that this procedure will come to an end on the seventh draw ?

Out of the 3 black balls 2 should have been drawn in the first six draws so that when the third appears in the seventh draw the procedure ends. In the first six draws 4 white balls and 2 black balls should  have been drawn so that the required probability
${=}^6{\mathrm{C}}_4\frac{10}{13}\cdot \frac{9}{12}\cdot \frac{8}{11}\cdot \frac{7}{10}\cdot \frac{3}{9}\cdot \frac{2}{8}\cdot \frac{1}{7}=\frac{15}{286}$