Probability of Drawing Two Balls of Different Color in Second Draw. Bob has eight balls in total; 3 are green, and 5 are red. Due to Bob's impulsive nature he spontaneously decides to play a game where he blindfolds himself and draw balls. Each draw consists of him taking two balls from the pool without returning the balls.
The probability of taking two balls of different colors is P(15/56). Probability of first ball taken being G(3/8) and R(5/7) for second ball taken. The order of taking the balls, whether Green before Red or Red before Green(hence becoming R(5/8) & G(3/7)) is interchangeable. Ultimately, the probability is as stated, P(15/56) for drawing balls of different color.
Bob is happy to know the probability because he enjoys color diversity. Unfortunately, Bob has extremely low self-esteem. Bob tends to doubt himself, thus causing him to discard the balls from his first draw. After discarding his first draw he will draw once more, and keep those balls regardless of result. 
Now he wants to figure out the probability of taking two balls of different colors in the second draw. Throughout this process Bob remains blindfolded, and he does not return the two balls that he's taken from the first draw(the pool is now 6).
Bob does not know the result of his first draw. How can he figure out the probability of taking two balls of different colors in his second draw?
The probability of drawing two balls of different colors for second draw would be P(1/6) if first draw results in two Green.
The probability of drawing two balls of different colors for second draw would be P(4/15) if first draw results in two different colors.
The probability of drawing two balls of different colors for second draw would be P(3/10) if first draw results in two Red.
As you can see all my answers are reliant on the initial draw. Is there an equation or something to determine the result without knowing answer to initial draw?
 A: As Paul noted the probability of two different balls in the first draw of two balls is ${15\over28}$.
In the given story we have a second draw of two balls, but we don't know what happened in the first draw. This just means that the second draw is a random draw of $2$ balls from $8$. This can be viewed in the following way: The $8$ balls are randomly put in a line, and we take balls $3$ and $4$ from the row. It is obvious that their being different (i.e., being green and red) is equally probable as the event that balls $1$ and $2$ are different, and this probability is ${15\over28}$. 
A: The probability of different color balls on the two draws is just the sum of the probabilities for the 2 events.  These 2 possibilities of having it are drawing a red first, and then a green, or a green first, and then a red.  Obviously both cannot happen simultaneously.  
The probability of green first and then a red is equal to $3/8 x 5/7=15/56$.  Going the other way yields $5/8 * 3/7=15/56$  Look at that, there are equal!  Think about why they should be equal.  So the total probability is $15/56+15/56=15/28$.
