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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.

Tony Stark
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    If I had the power, I'd delete every single question containing the word "intuition". –  Oct 22 '20 at 18:34
  • Nowhere in the problem does it say there are only $10$ balls in the bag. It says there are $10$ white balls and some number of black balls. – WaveX Oct 22 '20 at 18:40
  • @ProfessorVector i'm interested, why would you do so? – Alan Oct 22 '20 at 19:02
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    @Alan Intuition ("the power or faculty of attaining to direct knowledge or cognition without evident rational thought and inference" https://www.merriam-webster.com/dictionary/intuition) can't be explained by definition, it can only be acquired by exercise. Emphasis: by doing exercises, not by watching them done. –  Oct 23 '20 at 13:43
  • Does this answer your question? Intuitively, why does Bayes' theorem work? –  Dec 25 '21 at 22:46

1 Answers1

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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

  1. 4 w's : 1 element

  2. 3 w's : 4 elements

  3. 2 w's : 6 elements

  4. 1 w : 4 elements

  5. 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

$$\binom{3}{3} \binom{7}{3} ~\text{to} \binom{4}{3} \binom{6}{3} ~\text{to} \binom{5}{3} \binom{5}{3} ~\text{to} \binom{6}{3} \binom{4}{3} ~\text{to} \binom{7}{3} \binom{3}{3} $$

$$ = 35 ~\text{to} ~80 ~\text{to} ~100 ~\text{to} ~80 ~\text{to} ~35. $$

user2661923
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