Given a pdf $f_{Y}(y)$ and $n$ random observations. Find probability that last observation will be the smallest number in all the sample? Suppose that n observations are chosen at random from a continuous pdf fY(y). What is the probability that the last observation recorded will be the smallest number in the entire sample?
attempt: Suppose that $n$ observations are chosen at random from a continous pdf $f_{Y}(y)$.
THen the cdf is defined as $F_{Y}(y) = P(Y \leq y)$.
Then the probability that the last observation recorded will be the smallest number n the entire sample is
$P(y_{n} = y_{1}) = P[(y_{1} \leq y_{n}) \cap (y_{2} \leq y_{n}) \cap ....(y_{n-1} \leq y_{n})]$ =  $P[(y_{1} \leq y_{n})]P[(y_{2} \leq y_{n})]P[(y_{n-1} \leq y_{n})]$ = $[F_{Y}(y)]^{n-1}$
This problem is from the section ordered statistics. 
Is this a correct way? Please any feedback would be really appreciated. Thank you in advance.
 A: You have $n$ iid random variables, $\{Y_1, \ldots, Y_n\}$, and wish to know the probability that the $Y_n$ is less than all of the others.  You know the probability density function $f_Y$ and have found the cumulative distribution function $F_Y$.
You have almost got it.  What you have is the probability that a specific value of $Y_n$ is the greatest of the array.
What you need is to find the expectation of the probability that $Y_n$ has the least value of the array.
$$\mathsf P\left[\bigcap_{k=1}^{n-1}(Y_k > Y_n)\right] = \int_{\mathbb Y_n} \biggl(1-F_Y(y_n)\biggr)^{n-1} f_Y(y_n)\operatorname d y_n $$

Of course, there is a much more elegant solution.   If all variables are independently and identically distributed, then is there not an equal probability that any one of them will be the least ordered statistic?
$$\mathsf P\left[\bigcap_{k=1}^{n-1}(Y_k > Y_n)\right] = \frac 1 n$$
A: Continuous does not imply uniform. There's no reason to believe that each outcome is equally likely.
The problem does not state i.i.d. This is an assumption of one of the problem solvers and not the problem itself.
