Question

Probability,
A bag contains some white and some black balls, all combinations of balls being equally likely. The total number of balls in the bag is 10. If three balls are drawn at random without replacement and all of them are found to be black, the probability that the bag contains 1 white and 9 black balls is ...

Answer 1

@rational

Answer 2

looks hard

Answer 3

it is a standard question of bayes theorem but quite long...um looking for a shorter one

Answer 4

Let the number of black balss be b and number of white ballws be w. n+b=10.
The probability of this happening is bP3/(10P10). The probability of this configuration of balls is 1/11. So is this question asking P(9 black and one white|3 black is drawn)?
If three black is drawn, only 8 configuration of balls are valid as some don't have 3 balls. The probability of 3 black ball is drawn from the whole sample space is \(P(\text{3 black is drawn})=\sum_{k=3}^{10}kP3/10!\). The probability of having a 9 black and 1 white ball configuration is either 1/11 or 1/8. So\[
P(\text{9 black and one white|3 black is drawn})=\frac{\frac{1}{8}P(\text{3 black is drawn})}{P(\text{3 black is drawn})}\]

I don't know.

Answer 5

So P(9 black and one white|3 blacks is drawn) is 1/8? That kind of make sense as all configurations without 3 black balls are excluded and all other configurations are equally likely.

Answer 6

Could it be \[\frac{\frac{9P3}{10P3}}{\sum_{k=3}^{10}\frac{kP3}{10P3}}\]

Answer 7

That would be 14/55. Is that correct?

Answer 8

I am sure that it is not 1/8 as not all configurations of balls are equally likely.

Answer 9

NicexD

Answer 10

answer is correct but i was thinking as if to reduce the effort of finding P(3 black is drawn)
I noticed the denom is 11C2... so any guesses?