Probability that all black balls are drawn after the sixth draw.

Question

Consider a bag containing 10 balls of which a few are black balls. Probability that the bag contains exactly 3 black balls is 0.6 and probability that the bag contains exactly 1 black ball is 0.4. Now, balls are drawn from the bag, one at a time without replacement, till all black balls have been drawn. The probability procedure would end at the 6th draw is p, find 100p.

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My attempt: I assumed that the event bag contains 1 black ball (E) and the event bag contains 3 black balls (F) are mutually exclusive and exhaustive because P(E)+P(F)=1. Let A be the event of completely drawing all black balls in the bag in 6 draws

Therefore by Total Probability Theorem [(1)][1],

P(A) = P(E)*P(A|E) + P(F)*P(A|F). Where P(E)=0.4 and P(F)=0.6 (Given).

P(A|E)= $\frac{\binom{9}{5} .\binom{1}{1}}{\binom{10}{6}}$. Hence P(A|E)=$\frac{1}{6}$.

P(A|F)= $\frac{\binom{7}{3} .\binom{3}{3}}{\binom{10}{6}}$. Hence P(A|F)=$\frac{3}{5}$.

By applying total probability theorem, I get P(A)=$\frac{17} {50}$ and P(A)=p. Therefore 100p = 34, which is not even remotely close to the answer given in the booklet.

Answer:

9.

Links:
[1]: https://en.wikipedia.org/wiki/Law_of_total_probability

Answer 1

1. when there are 3 black balls and 7 other balls.

   the 6th ball taken must be the third black ball, hence in the first 5 picks there are three other balls and 2 black balls. also, the three black balls are to be treated as separate entities. no of favorable ways = $\binom{7}{3}.\binom{3}{2}.5!.1$ (order of picking objects is considered in probability)

2. when there is 1 black ball and 9 other balls.

   the 6th ball is the black ball, hence first 5 picks are other balls. no of favorable ways = $\binom{9}{5}.5!.1$

also total no of ways of picking 6 balls from 10 balls = $\binom{10}{6}.6!$

hence using total probability theorem, $$p = \dfrac{(0.4).\dbinom{9}{5}.5! + (0.6).\dbinom{7}{3}.3.5!}{\dbinom{10}{6}.6!} =\dfrac{9}{100}$$

Answer 2

You want the probability that the last black ball is drawn on the sixth draw -- this is that the remainder are placed among the first five draws, and none among the last four draws.

So the probability for event $A$ when given that $N$, the balls in the bag, equals $n$ is:

$$\begin{align}\mathsf P(A\mid N{=}n)~&=~\dfrac{\dbinom 5{n{-}1}\dbinom 11\dbinom 40}{\dbinom{10} n}\\[2ex]&=~\dfrac{\dbinom 5{n-1}}{\dbinom{10}n}\end{align}$$