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As the title explains, I never could understand probabilities. It's one of those things that how much I try, I can't quite understand.

I have to do one homework exercise about entropy and I'm given a set of probabilities. I know how to calculate entropy but I don't know how to interpret the given data.

The alphabet is $ S = \{1,2\}$ and the conditional probabilities are $P(1|1)=0.8\ P(2|1)=0.2\ P(1|2)=0.6\ P(2|2)=0.4$ and $P(1,2)=P(2,1)$

I've created this table (don't know if it is right or not):

       | X = 1 | X = 2
 Y = 1 |  0,8  |  0,2
 Y = 2 |  0,6  |  0,4

I know that I need the probability of $1$ and $2$ to calculate the entropy. To get the probability of $1$ is like this? $$P(1) = \frac{P(1,2)}{P(2|1)}$$

If so, How can I get $P(1,2)$?

I know that $P(1)=0.75$ and $P(2)=0.25$ but I don't understand how to get to this result

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up vote 1 down vote accepted

here, the explicit solution: You have, $$P(1) = \frac{P(1,2)}{P(2|1)}$$ and $$P(2) = \frac{P(2,1)}{P(1|2)}$$

Because of $$P(2,1) = P(1,2)$$ we get: $$P(1)*P(2|1)=P(2)* P(1|2)$$ $$P(1)=3*P(2)$$ Furthermore, you have $P(1)+P(2)=1$ and hence $$4*P(2)=1$$, Hence $$P(2)=0.25$$

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Hi. Thanks but I still don't understand. Don't I need at least $p(1)$ or $p2$? Another question. Is table correct? – Favolas Jan 16 '13 at 11:21
No, u do not need p(1) or p(2). You have to different equations with two variabels P(1) and P(2), so you can solve the system of equations. I do not know the conventions for the table, hence I judge its correctness. – user46128 Jan 16 '13 at 14:59
Hi. Once again thanks but still don't know how to solve it. I know that $P(1)=0.75$ and $P(2)=0.25$ but I don't understand how to get to this result – Favolas Jan 16 '13 at 15:46
Sorry for the late feedback. Thanks. That was so simple. Right in front of my eyes but the truth was that I never saw it. I told I don't understand probabilities :-) – Favolas Jan 16 '13 at 22:54

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