A bag contains 10 white and 3 black balls. Balls are drawn one by one without replacement till all the black balls are drawn. The probability that the procedure of drawing balls will come to an end at the 7t h draw is

Question

A bag contains 10 white and 3 black balls. Balls are drawn one by one without replacement till all the black balls are drawn. The probability that the procedure of drawing balls will come to an end at the 7t h draw is (A) (15/286) (B) (105/286) (C) (35/286) (D) (7/286)

Tardigrade Admin · Accepted Answer

The drawing procedure ends at 7t h draw, if 3r d black ball comes at 7t h draw. Hence, in 1s t six draws there will be 2 black balls and 4 white balls. So, required Probability =(10C4 × 3C2/13C6)× (1/7) =(15/286)

Q. A bag contains $10$ white and $3$ black balls. Balls are drawn one by one without replacement till all the black balls are drawn. The probability that the procedure of drawing balls will come to an end at the $7^{t h}$ draw is

$\frac{15}{286}$

$\frac{105}{286}$

$\frac{35}{286}$

$\frac{7}{286}$

The drawing procedure ends at $7^{t h}$ draw, if $3^{r d}$ black ball comes at $7^{t h}$ draw. Hence, in $1^{s t}$ six draws there will be $2$ black balls and $4$ white balls.

So, required Probability $= \frac{^{10} C _{4} \times _ ^{3} C _{2}}{^{13} C _{6}} \times \frac{1}{7}$ $= \frac{15}{286}$

Q. A bag contains 10 white and 3 black balls. Balls are drawn one by one without replacement till all the black balls are drawn. The probability that the procedure of drawing balls will come to an end at the 7th draw is

28615​

286105​

28635​

2867​