I am having a hard time developing an intuition behind Bayes' Theorem. I have asked this question with the help of the question below only because I was not able to get the right words to put in my question.
A bag contains 10 balls some of which are white and others are black. A person draws 6 balls and finds that 3 are black and the other 3 are white. Find the probability that the number of black balls in the bag are equal to the number of white balls.
I have already solved the question using Bayes' Theorem to get the answer as $10/33$. However I am confused as to why the solution below is invalid:
I already have 6 balls out from the bag. For satisfying my required condition, I need to have 2 black balls and 2 white balls in my bag.
Sample space $(S) =\{wwww,wwwb, wwbw, wbww, bwww, bbww, \ldots\}$
$$n(S)= 16 $$
$$n(E)= 6$$
$$P(E)=6/16$$
This solution seems to be invalid. However, I am not able to get a concrete logic as to why this happens. Please provide me with some arguments.
Best Answer
Since you are comfortable computing the actual answer $\frac{10}{33}$ for yourself, I can confine my response to an intuitive explanation of why your alternative analysis is invalid.
When you divided your sample space into 16 elements, the distribution was
4 w's : 1 element
3 w's : 4 elements
2 w's : 6 elements
1 w : 4 elements
0 w's : 1 element
You then (in effect) assumed that the proportional relationships between the 5 possible distributions is
$$ 1 ~\text{to} ~4 ~\text{to} ~6 ~\text{to} ~4 ~\text{to} ~1. $$
Given the constraints of the problem, and the required method in computing the accurate answer of $\frac{10}{33}$ the actual proportional relationships between the 5 possible distributions is
$$\binom{3}{3} \binom{7}{3} ~\text{to} \binom{4}{3} \binom{6}{3} ~\text{to} \binom{5}{3} \binom{5}{3} ~\text{to} \binom{6}{3} \binom{4}{3} ~\text{to} \binom{7}{3} \binom{3}{3} $$
$$ = 35 ~\text{to} ~80 ~\text{to} ~100 ~\text{to} ~80 ~\text{to} ~35. $$