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

Question

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

Tardigrade Admin · Accepted Answer

Required probability = text Probability of getting exactly two black balls and 4 text white balls in 1 text to 6 text th text draw × text probability of getting 3 text rd text black on 7 text th text draw =( 3 C 2 × 10 C 4/ 13 C 6) × (1/7)=(15/286)

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

$\frac{15}{286}$

$\frac{105}{286}$

$\frac{35}{286}$

$\frac{7}{286}$

Required probability $= Probability of getting exactly two black balls and 4 white balls in 1 to 6^{th} draw$
$\times probability of getting 3^{rd} black on 7^{th} draw$
$= \frac{^{3} C _{2} \times ^{10} C _{4}}{^{13} C _{6}} \times \frac{1}{7} = \frac{15}{286}$

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

28615​

286105​

28635​

2867​

Required probability = Probability of getting exactly two black balls and 4 white balls in 1 to 6th draw × probability of getting 3rd black on 7th draw =13C6​3C2​×10C4​​×71​=28615​

$\frac{15}{286}$

$\frac{105}{286}$

$\frac{35}{286}$

$\frac{7}{286}$

Required probability $= Probability of getting exactly two black balls and 4 white balls in 1 to 6^{th} draw$
$\times probability of getting 3^{rd} black on 7^{th} draw$
$= \frac{^{3} C _{2} \times ^{10} C _{4}}{^{13} C _{6}} \times \frac{1}{7} = \frac{15}{286}$