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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

Probability - Part 2

Solution:

Required probability $=\text { Probability of getting exactly two black balls and } 4 \text { white balls in } 1 \text { to } 6^{\text {th }} \text { draw } $
$\times \text { probability of getting } 3^{\text {rd }} \text { black on } 7^{\text {th }} \text { draw } $
$=\frac{{ }^3 C _2 \times{ }^{10} C _4}{{ }^{13} C _6} \times \frac{1}{7}=\frac{15}{286} $