You are on the right track , though that is not the Probability we want.
What we want is Conditional Probability like this :
$$P(E) = \frac{P(BBB) + P(BRB) + P(RBB)}{P(BBB) + P(BRB) + P(RBB) + P(BBR) + P(BRR) + P(RBR)}$$
[[ $E$ is the "Event" concerned : third Ball Blue , given at least 1 Blue Ball among first & second ]]
We can then make it :
$$P(E) = \frac{P(BBB) + P(BRB) + P(RBB)}{P(BBX) + P(BRX) + P(RBX)}$$
[[ $X$ indicates either ]]
Hence it is just :
$$P(E) = \frac{P(BBB) + P(BRB) + P(RBB)}{P(BB) + P(BR) + P(RB)}$$
ADDENDUM :
(1) Probability OP gave will match the Event "take 3 Balls , at least 3 Balls are Blue , last Ball is Blue" , which is not Conditional Probability
(2) When we consider all Events , Probability should add up to Exactly $1.0$ always , where the Solution given here will work out : Add that Probability to the Event that the last Ball is Red & we will indeed get $1.0$ Consistently.
The value given by OP will add up to very tiny total , which will alert us that something is wrong.
(3) When Events (like taking 2 Balls out) have already occurred , the Sample Space will change accordingly , incorporating those states.
We have to work within the new Sample Space.
That is Conditional Probability in a nut-shell.