Understanding the distribution of the minimum of two exponential random variables

I looked at the comments of this question and it had confused me a little: How to evaluate probability of minimum and maximum of three random variable.

My Problem: Let $T_A,T_B, T_C$ are the i.i.d random variables that follows the Exponential distribution Exp(1) i.e. parameter $λ=1$ . I want to find $P(T_A < min(T_B,T_C))$

My understanding is that the minimum is another exponential random variable with rate (1+1) in this case.

So $P(T_A < min(T_B,T_C)) = \frac{1}{1+1+1} = \frac{1}{3}$

The comment which I am referring to mentioned :

$P(T_A < min(T_B,T_C))=P(T_A<T_B)P(T_A<T_C) = \frac{1}{2}.\frac{1}{2}=\frac{1}{4}$

Which one is correct?

Since $T_A, T_B, T_C$ are iid exponential distributions, they must be observed in one of the following orders, each of which is equally likely with probability $1/6$. We omit the case where two or more are equal since this occurs with probability zero:

$$T_A < T_B < T_C \\ T_A < T_C < T_B \\ T_B < T_A < T_C \\ T_B < T_C < T_A \\ T_C < T_A < T_B \\ T_C < T_B < T_A$$

In only the first two cases is $T_A < \min(T_B, T_C)$ satisfied.

The claim $$\Pr[T_A < \min(T_B, T_C)] = \Pr[T_A < T_B]\Pr[T_A < T_C]$$ is incorrect, because the event $T_A < T_B$ includes the case $T_C < T_A < T_B$, and the event $T_A < T_C$ includes the case $T_B < T_A < T_C$. So although it is true that

$$T_A < \min(T_B, T_C) \equiv (T_A < T_B) \cap (T_A < T_C),$$

it is not true that $$\Pr[(T_A < T_B) \cap (T_A < T_C)] = \Pr[T_A < T_B]\Pr[T_A < T_C]$$ because the events $T_A < T_B$ and $T_A < T_C$ are not independent. You can see this explicitly from the table above: $T_A < T_B$ includes the first, second, and fifth inequalities, of which only the first two also satisfies $T_A < T_C$. So $$\Pr[T_A < T_C \mid T_A < T_B] = 2/3 \ne \Pr[T_A < T_C] = 1/2.$$