0
$\begingroup$

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?

Thanks for your assistance!

$\endgroup$
0
$\begingroup$

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 $D \neq B$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.