$\because$   A = {1, 0, –1, 2}

(a) For skew symmetric matrix, all diagonal elements can be filled in only one way by zero.

For a_ij = – a_ji

$\therefore$ For every a_ij(i $\ne$j)  there are three ways, then for corresponding a_ji there is only way

$\therefore$  Required probability = $\frac{3\times 3\times 3}{4^9}=\frac{27}{4^9}$

(b) For matrix to be symmetric, a_ij = a_ji

$\therefore$ Probability

$=\frac{4\times 4\times 4\times 4\times 4\times 4\times 1\times 1\times 1}{4^9}=\frac{1}{64}$

(c) Total number of function = 4⁴

number of onto functions = 4!

$\therefore \mathrm{Probability}=\frac{4!}{4^4}=\frac{24}{256}=\frac{3}{32}$

(d)  f (i) $\ne$ i

$\therefore \mathrm{number}\mathrm{of}\mathrm{function}=4!(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!})=9$

$\therefore \mathrm{Probility}=\frac{9}{4!}=\frac{9}{24}=\frac{3}{8}$