• # question_answer A bag contains 10 black and 3 white balls. Balls are drawn one-by-one with out replacement till all the white balls are drawn. The probability that the procedure of drawing balls will come to an end at the seventh draw is A)  $\frac{15}{286}$                  B)  $\frac{7}{286}$ C)  $\frac{105}{286}$                            D)  $\frac{35}{286}$

Required probability = Probability of getting exactly two white balls and four black balls is 1 to 6th draw(X) probability of getting third white ball in 7th draw. $=\frac{^{3}{{C}_{2}}{{\times }^{10}}{{C}_{4}}}{^{13}{{C}_{6}}}\times \frac{1}{7}=\frac{15}{286}$