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I am given a question:

A bag contains 4 Balls. Two balls are drawn at random and found to be red. What is the probability that all balls are red?

what I have done:

enter image description here

using Bayes Theorem I have calculated the Required probability by this formula:

$$\sf P(E_3\mid E) ~=~ \dfrac{P(E\mid E_3)\cdot P(E_3)}{P(E\mid E_3)\cdot P(E_3) ~+~ P(E\mid E_2)\cdot P(E_2) ~+~ P(E\mid E_1)\cdot P(E_1)}$$

but the answer doesn't seem to match, where I have gone wrong?

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    I don't know if this is a Bayes theorem question. but I think I have applied the correct method. I may be wrong. – 2017-01-17
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    Why should we take as a prior the belief that those three cases are equiprobable? That seems very unlikely. I have no idea what the right prior should be...is there some missing information? – 2017-01-17
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    Just to stress...if I said I drew two balls out and observed that they both had mass then would you apply this "equiprobable prior" to the mass of the other balls? Presumably not, since our prior here is that "all balls have mass". To build a prior in your case would seem to require me to know something about what percentage of balls are red, and I don't. – 2017-01-17
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    [link](https://www.youtube.com/watch?v=Zxm4Xxvzohk) check this out, They really explain well – 2017-01-18
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    @lulu The question is correct it was asked in my examination and I found it on one book also but the solution was not given. now there are 3 possibilities: i,e. having (2,2), (3,1) and (4,0) and all three are equally likely to happen.so I took '1/3', am I wrong? – 2017-01-18
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    @JayanthKumar that was really helpful, great. – 2017-01-18
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    I see no justification for assuming those three state are equiprobable. If you imagine that each ball is Red with probability $p$ then, a prior, $(2,2)$ has probability $\binom 42 p^2(1-p)^2$, $(3,1)$ has probability $\binom 43 p^3(1-p)$, while $(4,0)$ has probability $p^4$. No value of $p$ makes all three equal. One could declare that an external force chose equally between these options...but that's a bizarre assumption. – 2017-01-18
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    Look at my (caricature) example: "In your situation I draw two balls and observe that both have mass. What is the probability that all the balls have mass?" Here the answer is $1$ because my (very strong) prior is that all balls have mass. Would you really reply that all three states had the same probability? – 2017-01-18
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    Oh okay, I am a high school student and I do not know from where those complex probabilities you made came from but I did what I learned in school and my teacher taught me. seems I need to study more to get those concepts. – 2017-01-18

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Let $E$ be the event that two red balls are drawn. $E_1$ is the scenario where there are two red balls and two other. The probabilty of drawing two red balls is in this situation is $1/6,$ so we have $$ P(E|E_1) = 1/6.$$

In your picture you have $2/4,$ so it seems you've gone wrong in calculating the conditional probabilities. (The Bayes' formula you wrote down looks correct, although I'm not sure if your priors of $P(E_1) = P(E_2) = P(E_3) = 1/3$ are correct... were you asked to assume that?)

To understand why $1/6$ is correct, recall that there are $6 = {4\choose 2}$ ways to pick two balls. Only in one of the ways is it the case that both balls are red.

$P(E|E_2)$ needs to be recalculated as well. $P(E|E_3) = 1$ is correct. If all four balls are red, you will draw two red balls every time.

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    there are 3 possibilities: i,e. having (2,2), (3,1) and (4,0) and all three are equally likely to happen.so I took '1/3' – 2017-01-18
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    okay, now I figured out my mistake, I am required to apply combinations in all three cases while calculating P(E/En) and the answer seems to be correct now. success. – 2017-01-18