Consider a bag containing 10 balls of which a few are black balls. Probability that bag contains exactly 3 black balls is 0.6 and probability of bag containing exactly 1 black ball is 0.4. Now, balls are drawn from the bag, one at a time, with out replacement, till all black balls have been drawn.Find the probability that this procedure would end at the `6^(th)` draw.

Explanation

given that n= 10

let the probability of event occuring be P

then, `P(6) = ((0.6)* (3*2*1 * 7*6*5*1))/ (10*9*8*7*6*5) * 5!`

`P(6) = (0.6)`

`P(6) = (0.4)*(9*8*7*6*5)/(10*9*8*7*6*5)`

=`(0.4)/10 = 1/25`

`p(6)= 0.6 + 1/25`

= `3/5 + 1/25 = (15+1)/25 = 16/25`