As $0.4+0.6=1$, the man either takes a step forward or a step backward. Let a step forward be a success and a step backward be a failure. Then, the probability of success in one step

$=p=0.4=\frac{2}{5}$

The probability of failure in one step

$=\mathrm{q}=0.6=\frac{3}{5}$

In 11 steps he will be one step away from the starting point if the numbers of successes and failures differ by 1 . So, the number of successes $=6$

The number of failures $=5$

or the number of successes $=5$, The number of failures $=6$

$\therefore$ the required probability

$={ }^{11} \mathrm{C}_{6} \mathrm{p}^{6} \mathrm{q}^{5}+{ }^{11} \mathrm{C}_{5} \mathrm{p}^{5} \mathrm{q}^{6}$

$={ }^{11} \mathrm{C}_{6}\left(\frac{2}{5}\right)^{6} \cdot\left(\frac{3}{5}\right)^{5}+{ }^{11} \mathrm{C}_{5}\left(\frac{2}{5}\right)^{5}\left(\frac{3}{5}\right)^{6}$

$=\frac{11 !}{6 ! 5 !}\left(\frac{2}{5}\right)^{5} \cdot\left(\frac{3}{5}\right)^{5}\left\{\frac{2}{5}+\frac{3}{5}\right\}$

$=\frac{11.10 .9 .8 .7}{120} \cdot \frac{2^{5} \cdot 3^{5}}{5^{10}}=462 \times\left(\frac{6}{25}\right)^{5}$