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Let's say I have a basket with green and red balls. $70\%$ of the balls are green (let's call the fraction $f_g = 0.7$), the rest is red - $f_r=0.3$.

Now, suppose I take one ball from the basket and guess its color (without looking, of course). What's the probability that my guess was wrong?

Supposedly the answer is $f_g(1-f_g)+ f_r(1-f_r)$, but I cannot convince myself this is correct. How to describe the problem mathematically (sample space, etc.)?

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  • $\begingroup$ I agree that $f_g$ is the probability of picking a green ball ($f_r$ - red). But why multiply it by $1-f_g$? How do I know that these events independent and I can multiply them? $\endgroup$
    – user4205580
    May 7, 2015 at 12:04
  • $\begingroup$ Sorry, I didn't read carefully enough. What are the probabilities of your guesses? $\endgroup$
    – David Mitra
    May 7, 2015 at 12:08
  • $\begingroup$ Do you guess randomly? I mean is your guess "blue" in 50% of times and red in other 50% of times? or knowing the fractions change the probability of your guesses? $\endgroup$
    – Ebrahim Ghasemi
    May 7, 2015 at 12:11
  • $\begingroup$ I think the answer depends on the probabilities of you guessing green and guessing red. You haven't specified them. The given answer suggests that you guess, randomly, green 70% of the time and red 30% of the time. $\endgroup$
    – Mick A
    May 7, 2015 at 12:12
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    $\begingroup$ @Abraham The min probability of guessing wrong would be when you guess green every time (because most balls are green), giving P(Wrong)=0.3. $\endgroup$
    – Mick A
    May 7, 2015 at 12:25

2 Answers 2

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There are two situations :

  1. You guess the color really randomly, i.e. you guess it is green for $50$% of times and you guess it is red for the other $50$% of times.
  2. You guess efficiently to reduce the number of wrong guesses. In this situation, you must check to is if you guess green for $70$% of times and guess red for $30$% of times it is better or if you guess the color that we have more.

For both situations we have :

P(Wrong guess) = P(Guess is Green & You pick Red) + P(Guess is Red & You pick Green)

As your guess and the pick are independent, the probability of each part is multiply of its parameters, i.e. :

$P(Wrong_{Guess}) = P(Guess_{Green})* P(Pick_{Red})+P(Guess_{Red})*P(Pick_{Green})$

So for 1st situation :

$P(Wrong_{Guess})= 1/2 * 0.3 + 1/2* 0.7 = 0.5$

And for 2nd situation :

$P(Wrong_{Guess})= 0.7 * 0.3+ 0.3*0.7= 0.42 $

Look, $42$ % is greater than $30$%. So the best way to minimum the wrong guesses is to say Green for all the times, And in this situation, $P(Wrong_{Guess}) = 0.30$

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If 'guessing' means that there is a chance of $p_r$ that you guess red and a chance of $p_g$ (here $p_r+p_g=1$) that you guess green then the probability that your guess is wrong is: $$f_g\times p_r+f_r\times p_g$$

'Without looking' stands for independence of the two probabilities: the first concerns taking a ball, the second concerns making a guess.

It is up to yourself (as the one who guesses) how $\langle p_r,p_g\rangle$ looks like.

  • In special case $p_r=p_g=\frac12$ this equals $f_g\times \frac12+f_r\times\frac12=0.5$.
  • In special case $p_r=1\wedge p_g=0$ this equals $f_g=0.7$.
  • In special case $p_r=0\wedge p_g=1$ this equals $f_r=0.3$.
  • In special case $p_r=f_r\wedge p_g=f_g$ this equals $2f_gf_r=0.42$.
  • In special case $p_r=f_g\wedge p_g=f_r$ this equals $f_g^2+f_r^2=0.58$.
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  • $\begingroup$ I always thought the independence is just a mathematical concept, that says events $A$ and $B$ are independent iff $P(A \cap B)=P(A)P(B)$. We can't just guess that it holds by reading the description of our problem. Rather, we should prove it holds. I guess sometimes two events are SEEMINGLY independnet, and then it turns out they aren't. Or I don't understand independence... $\endgroup$
    – user4205580
    May 7, 2015 at 12:24
  • $\begingroup$ The description of our problem provides the inspiration to make a mathematical model of the described reality. Based on the description (and meeting what you call seemingly independence) we can conclude that independence is on its place as a part of that model. $\endgroup$
    – drhab
    May 7, 2015 at 12:34
  • $\begingroup$ So we need to take the fact that those events are independent as an axiom? We can't prove it, just like the validity of mathematical model of any informally stated problem in probability or any other mathematical theory. $\endgroup$
    – user4205580
    May 7, 2015 at 12:35
  • $\begingroup$ Axioms are part of the theory and not assumptions that are made based on the description. Btw, do you 'believe' in independence here? If so then how would you prove it without preassuming and without referring to 'without looking'? $\endgroup$
    – drhab
    May 7, 2015 at 12:46
  • $\begingroup$ Yes, I should've used the word 'assumption', not 'axiom. We cannot prove it. As with every such problem in probability we do our best to come up with a 'reasonable' probabilistic model, but we cannot be sure if it's valid. Do you agree with me? $\endgroup$
    – user4205580
    May 7, 2015 at 12:55

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