Guessing the color of a ball from the basket 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.)?
 A: There are two situations : 


*

*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.

*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, $P(Wrong_{Guess}) = 0.30$
A: 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 $f_g^2+f_r^2=0.58$.

