Bag contains $6$ balls. It is not known what colors they are. 3 balls are drawn from the bag and found to be black....

Question

A bag contains $6$ balls, and it is not known what colors they are. $3$ balls are drawn from the bag, and found to be black. Find the probability that no black balls are left in the bag now.

I'm stuck on this problem. Initially I made cases if $3$ are picked, then the options for the number of black balls remaining would be $\{0,1,2,3\}$.

Or rather, another way to put it could be- 'Possibility of total no of black balls that can be in the bag'={3,4,5,6}

Please assume all numbers of black balls in the bag to be equally likely, i.e that the total number of possibilities is equally likely.

There's only one case, $\{0\}$, where the remaining can be zero?

Thanks in advance.

Answer 1

I must say that the question has insufficient information. We have to put in mind that there is an unlimited number of possible colors. The possible color of the ball is not limited that's why we can't calculate the probability. Here is what I mean.

When calculating probability we have to put into mind all possibilities then add the factors that will affect each possible event. If the first three balls were found to black the probability of the next ball to be black is $\frac{1}{∞}$. That's because there is an unlimited number of colors depending on logical and wavelength distinction capability of the eye.

You may argue that the eye can interpret visible light on a limited range but I can use an analogy to justify my statement. A typical human eye will respond to wavelengths from about 380 to 740 nanometers. Suppose it was a number line, we know that there are infinite number of decimals that can be inserted. Example; 380, 380.1, 380.01 etc., and each corresponds to a slightly different color. With this we can conclude that the question is not answerable.

Answer 2

The problem as quoted is ill-posed, as no distribution is specified. You subsequently specify a distribution, but this distribution makes the problem trivial: If all numbers of black balls consistent with the observations are equiprobable, then, since there are $4$ such numbers, the probability for each of them is $\frac14$.

If each ball in the bag was initially equally likely to be black or not, the observation of $3$ black balls eliminates

$$ \binom60+\binom61+\binom62=1+6+15=22 $$

of the $2^6=64$ a priori possibilities. The possibility that there are no black balls left accounts for $\binom63=20$ of the remaining $64-22=42$ possibilities, so in this scenario the probability that no black balls are left would be

$$ \frac{20}{42}=\frac{10}{21}\approx47.6\%\;. $$