GRE Probability-Total Probability & Bayesian 
Six billiard balls, numbered 1 through 6, are placed in a box. Three of the balls are red, and three of the balls are blue. One ball is to be randomly drawn from the box.

Quantity A
The probability that the ball will be an even numbered red ball
Quantity B
0.5
Answer Options (The Answer is D)
A) Quantity A> Quantity B 
  B) Quantity B> Quantity A 
  C) Quantity A=Quantity B  
  D) Cannot be Determined   
Approach
To find: P(Even/Red)
i. P(Red)=0.5 
ii.P(Red && Even) can be 0, 0.16, 0.33 OR 0.5 
Answer: SUM( P(Red && Even))/ P (Red)  { Basically , Sum of P( 'i' even balls/red) where i ranges from 0 to 3 }
    = (0+0.16+0.33+0.5)/0.5

This is greater than 1, which is an erroneous value.
Is the use of Total Probability valid in this case at all? I am unable to understand when to apply the OR concept in cases when calculating probability , as in this case.
Thanks for the help!
 A: The answer can't be determined as it stands...that is, without having more information or making some assumptions.  It is possible that the probability is $\frac 12$ (if all the even balls are red), but it is also possible that the probability is $<\frac 12$ (any other distribution of colors).
One natural assumption you might make:  assume that the balls were painted by extracting three balls at random (without replacement) and painting those red, and the others blue. This is the "unbiased" assumption, in which the colors are independent of the parities.  The answer to your question, under this assumption, should therefore be $\frac 12\times \frac 12=\frac 14$. Let's see in detail how that plays out: (here $O$ means "odd" and $E$ means "even")
The probability that you get $OOO$ equals the probability of $EEE$ equals $$\frac 12\times \frac 25 \times \frac 14=\frac 1{20}$$
The probability of drawing, say, $EOO$ in that order is $\frac 12\times \frac 35 \times \frac 12=\frac 3{20}$  from which it follows that the probability of getting exactly one $E$ or exactly one $O$ is $$\frac 9{20}$$
Inspection shows that these are all the cases...that is, either all three have the same parity or exactly one is of a given parity. Indeed, $\frac 1{20}+\frac 2{20}+\frac 9{20}+\frac 9{20}=1$
It is easy to read off the conditional probabilities (here the event $X$ is the one you seek..."the selected ball is even and red":
$$P(X|OOO)=0\;\;\;\;P(X|EEE)=\frac 12\;\;\;\;P(X|E00)=\frac 16\;\;\;\;P(X|OEE)=\frac 26$$
Where (just to be clear) a state like $EOO$ just means that exactly one of the red  balls is even (no condition on the order in which the balls were drawn).
We conclude with $$P(X)=\sum_{states}P(X|S)P(S)=0\times \frac 1{20}+\frac 12\times \frac 1{20}+\frac 16\times \frac 9{20}+\frac 26\times \frac 9{20}=\frac 14$$
Just to stress:  It is not clear to me that we are intended to make this assumption.
Note:  you can say that $A≤B$ (though that is not one of the stated options).  Indeed, that statement is true for all distributions of color so no extra assumptions are required.
