I am having trouble understanding how to begin with this question:
If ten points are picked independently and at random on the interval (0,1):
(a) What is the probability that point nearest 1 (i.e., the largest of the 10 numbers selected) will lie between 0.9 and 1.0.
(b) The probability is 1/2 that the point nearest 0 will exceed what number?
I'm assuming I need to apply the formula below somehow with $n = 10$ and $k = 10$ (since this is the largest value in the sample) and assuming $X$ follows a uniform distribution.
$$ g(x_{(k)}) = nF^{n-1}(x_{(n)})f(x_{(n)}) = \left\{\begin{array}{cl} nx^{n-1}_{(n)} & \text{if}\ 0 < x_{(n)} < 1;\\ 0 & \text{otherwise}. \end{array}\right. $$
Call $X_1,\dots,X_n$ the random sample, and $A=\min_i(X_i), B=\max_i(X_i)$.
Then, for $t\in[0,1]$,
$$P(B\le t)=P(\cap \{X_i\le t\})=\prod_iP(X_i\le t)=t^{10}$$
Hence
$$P(0.9\le B\le 1)=1-0.9^{10}\simeq0.65$$
Likewise,
$$P(A\ge t)=P(\cap \{X_i\ge t\})=\prod_iP(X_i\ge t)=(1-t)^{10}$$
Therefore $P(A\ge t)=\frac12$ iff $(1-t)^{10}=\frac12$, iff $t=1-2^{-1/10}\simeq 0.067$