Even with your updated version, there is not a well defined answer (which does not make this a bad question!) You need to have a prior distribution on the probability that the box contained various numbers of balls to start with.

Let's do some small examples. Suppose, that you are given a box with three balls. You sample one ball, and it is white. What should your estimate be that the majority of the balls are white? As we will see, we can't answer this question without some prior beliefs about how the balls were placed in the box.

Suppose, at first, that you know the two balls were chosen at random from a large collection, with equally many white and black balls. So, before sampling, your probability estimates were $1/8$ for three black balls, $3/8$ for BBW, $3/8$ for BWW, and $1/8$ for WWW. The probability that the initial distribution was BBW, and that you would sample a white, is $(3/8)(1/3)=1/8$. The corresponding probabilities for BWW and WWW are $1/4$ and $1/8$. So, conditioned on the fact that you sampled a white, the probability that the balls
are BBW is $(1/8)/(1/8 + 1/4 + 1/8) = 1/4$, and we see that the probability that the majority are white is $3/4$.

But now suppose that white balls are very rare. For example, say your box was packed from a collection which had 9 black balls for every 1 white ball. Then the corresponding probabilities before sampling are $(9/10)^3$ for BBB, $3\*(1/10)\*(9/10)^2$ for BBW, $3\*(1/10)^2\*(9/10)$ for BWW and $(1/10)^3$ for WWW. The probabilities that these distributions would hold and that you would sample a white ball are $(1/10)\*(9/10)^2$ for BBW, $2\*(1/10)^2\*(9/10)$ for BWW and $(1/10)^3$ for WWW. So, conditioned on you sampling a white ball, the probability that the box holds BBW is
$$\frac{(1/10)\*(9/10)^2}{(1/10)\*(9/10)^2+ 2\*(1/10)^2\*(9/10)+(1/10)^3},$$
and so forth. I'm too lazy to calculate the probability that the box is majority white in this case, but it should be much less than $3/4$.

This is why polls always report their margin of error as a $p$-value. When a pollster says "75% of Americans will vote for Kodos, with a margin of error of $\pm 10%$", this means "the probability that I would have obtained these poll results, conditioned on the assumption that either fewer than 65% or more than 85% of Americans will vote for Kodos, is $<.05$." (The use of $.05$ as a threshold is traditional, and has no deep significance.) If you have prior reason to know that the majority is very likely to vote for Kang then, even after you see the poll, the odds of Kodos winning might still be slim.