Answer
a)$\frac{1}{2}$
b)$\frac{1}{2}$
c)$\frac{1}{n}$
d)$\frac{1}{4}$
e)$\frac{1}{3}$
Work Step by Step
a) In any permutation one of them will precede the other. Since the two events of $1$ preceding $2$ and $2$ preceding $1$ are equally likely, hence $P(1 precedes 2) =\frac{1}{2}$.
b) is some us part a), i.e., $\frac{1}{2}$
c)Choose any two consecutive places out of n places in $(n-1)$ ways and put them accordingly, i.e., $1$ before $2$. The rest of the numbers can permute in $(n -2)!$ ways. So the probability is $\frac{(n-1)(n-2)!}{n!} =\frac{1}{n}$.
d) Both the probabilities are $\frac{1}{2}$ following the same logic in a) and since the two events are independent, hence their combined probability is $(\frac{1}{2})^2=\frac{1}{4}$
e)Any three places can be chosen in $^nC_3$ ways, and for each of these ways, $1$ and $2$ can interchange their positions. Rest of the numbers can be arranged in $(n-3)!$ ways. So the probability is $\frac{2^nC_3(n-3)!}{n!}=\frac{1}{3}$
