A man takes a step forward with probability 0.4 and backward with...

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 $={ }^{11} \mathrm{C}_{6} \mathrmp^{6} \mathrmq^{5}+{ }^{11} \mathrm{C}_{5} \mathrmp^{5} \mathrmq^{6}={ }^{11} \mathrm{C}_{6}\left(\frac{2}{5}\right)^{6} \cdot\left(\frac{3}{5}\right)^{5}+{ }^{11} \mathrm{C}_{5}\left(\frac{2}{5}\right)^{5}\left(\frac{3}{5}\right)^{6}=\frac{11 !}{6 ! 5 !}$\left(\frac{2}{5}\right)^{5} \cdot$\left(\frac{3}{5}\right)^{5}\left\{\frac{2}{5}+\frac{3}{5}\right\}=\frac{11.10.9.8.7}{120} \cdot \frac{2^{5} \cdot 3^{5}}{5^{10}}=462 \times\left(\frac{6}{25}\right)^{5}$