# Conditional Probability involving a mathematical sequence

I have a sequence of elements:

$$T_1 \hspace{1cm} P(T1 = A) = .5 \, , \\ T_2 \hspace{1cm} P(T2 = B) = .2\, , \\ T_3 \hspace{1cm} P(T3 = C) = .3\, .$$

Given the sequence $$TT = (T_1 = A, T_2 = B, T_3 = C)$$, the probability of that specific sequence will be $$.5 \cdot .2 \cdot .3 = .03$$.

But what is the conditional probability of $$P(T_2 | TT)$$ ?

Since we know that $$TT$$ is $$(A, B, C)$$, it seems that the conditional probability $$P(T_2 = B | TT) = 1$$.

It that correct?

First, in general $$P(TT)=P(T_1=A, T_2=B, T_3=C)= \\ P(T_1=A) \cdot P(T_2=B) \cdot P(T_3=C) = .5 \cdot .2 \cdot .3$$ only if $$T_1$$, $$T_2$$ and $$T_3$$ are three mutually independent variables. If they are not, the probability would be $$P(TT)=P(T_1=A, T_2=B, T_3=C)= \\ P(T_1=A | T_2=B, T_3=C) \cdot P(T_2=B |T_3=C) \cdot P(T_3=C)$$
Secondly, indeed $$P(T_2=B|TT) = P(T_2=B| T_1=A, T_2=B, T_3=C) =1$$, as the condition ensures that $$T_2=B$$, regardless of whether $$T_1$$, $$T_2$$ and $$T_3$$ are mutually independent variables or not.
We also have that $$P(T_2=D|TT)=0$$ for all $$D \neq B$$.