Answer:

Let A be the event that you pick the box with two red balls,

let B be the event that you pick the box with two blue balls, and

let C the event that you pick the box with one of each.

Let R be the event that you choose a red ball.

Then P(A)=P(B)=P(C)=13 .

Also P(R|A)=1 , P(R|B)=0 , and P(R|C)=12 .

Also P(R)=1×13+0×13+12×13=12 (law of total probabilities).

Then P(A|R)=P(A)P(R|A)P(R)=23 ,

P(B|R)=P(B)P(R|B)P(R)=0 , and

P(C|R)=P(C)P(R|C)P(R)=13 .

I presume you want the conditional probability. The conditional probability that the other ball is red given that a red was drawn is 23×1+0×0+13×0=23 .

Of course the unconditional probability is 12 because picking one ball is the same as not picking the other. So if one is 12 , the other is 1−12=12 .