100 numbers chosen on unit interval 100 numbers are chosen from the interval [0, 1], independently of each other. What is the probability that the 2nd largest # is <1/2? My reasoning was a) all 100 chosen are less then 1/2 
b) 99 are less then 1/2 and one is more 
This would give (1/2)^100 OR the probability of 99 less then 1/2 and one greater . Would i not need binomial theorem since im also factoring the number if ways to CHOOSE the 99 numbers ?? 

I would go about it using order statistics. Let the 100 independent random variables are $X_1,\,X_2,\,\cdots,\,X_{100}$ where all are uniformly distributed in $[0,\,1]$. Arrange them as $$X_{(1)}\leq X_{(2)}\leq\cdots\leq X_{(99)}\leq X_{(100)}$$ Then you need to find $$Pr\left[X_{(99)}<\frac{1}{2}\right]=\sum_{i=99}^{100}{100\choose i}F_X^i(0.5)\left[1-F_X(0.5)\right]^{100-i}$$ where $F_X(x)$ is the CDF of the random variables.
