Let $\tau_1, \tau_2, ..., \tau_K$ be i.i.d. exponential random variables with distribution $P(\tau_k<t) = 1 - e^{-\lambda t}$. Let $\tau^*_i$ be the $i^{th}$ order statistic. The p.d.f. of $\tau^*_i$ can be found in standard textbook: $$ f_{\tau_i^*}(t) = K{K-1 \choose i-1} f_{\tau}(t) (F_{\tau}(t))^{i-1}(1-F_{\tau}(t))^{k-i}$$
In my previous lecture, my professor was interested in computing the probability: $$ P(\tau_1^* = \tau_j \in [a,b))$$ for every $i \in \{1,...,K\}$, where $a,b \in \mathbb{R_+}$.
There is clearly some abuse of notation here since the event ${\tau_1^* = \tau_j}$ never happens almost surely. However he claims this notation is intuitive in some sense. He then writes that: $$ P(\tau_1^* = \tau_j \in [a,b)) = \int_{a}^{b} \lambda_j e^{-\lambda_j t} \prod_{i \neq j} e^{-\lambda_i t} dt$$
and if we have i.i.d. condition then:
$$ P(\tau_1^* = \tau_j \in [a,b)) = \int_{a}^{b} \lambda e^{-K \lambda t} dt$$ If we use the standard result from above for the p.d.f. of $\tau_1^*$, we get:
$$ f_{\tau_1^*}(t) = {K \choose 1} \lambda e^{-\lambda t} (1 - e^{-\lambda t})(e^{-\lambda t})^{K-1}$$ So we should have that: $$P(\tau_1^*\in [a,b))=\int_{a}^{b}{K \choose 1} \lambda e^{-\lambda t} (1 - e^{-\lambda t})(e^{-\lambda t})^{K-1}dt = \int_{a}^{b}K\lambda e^{-K\lambda t} (1 - e^{-\lambda t})dt$$
Why are these probabilities not equal? When considering the first order statistic, what does the event $\{\tau_1^* = \tau_j \in [a,b)\}$ mean?
The event in question is the event that the $j^{\rm th}$ sample point is the minimum and that it happens to fall in the interval $[a,b)$. By symmetry, the probability of this event is $$ \eqalign{ {1\over K}P(\tau_1^*\in[a,b)) &={1\over K}\int_a^b Kf_\tau(t)[1-F_\tau(t)]^{K-1}\,dt\cr &={1\over K}\int_a^b K \lambda e^{-\lambda t}[e^{-\lambda t}]^{K-1}\,dt\cr &=\int_a^b \lambda e^{-\lambda Kt}\,dt\cr &={e^{-\lambda Ka}-e^{-\lambda Kb}\over K}.\cr } $$ This answer is the same as would be found by using your instructor's method.