A man takes a step forward with probability 0.4 and backward with probability 0.6. The probability that at the end of eleven steps, he is just one step away from the starting point is

$$\begin{gathered} \mathrm{Let} \mathrm{a} \mathrm{step} \mathrm{forward} \mathrm{be} \mathrm{a} \mathrm{success} \mathrm{and} \mathrm{a} \mathrm{step} \mathrm{backward} \mathrm{be} \mathrm{a} \mathrm{failure}. \\ \mathrm{Then}, \mathrm{the} \mathrm{probability} \mathrm{of} \mathrm{success} \mathrm{in} \mathrm{one} \mathrm{step} =\mathrm{p}=0.4,\mathrm{n}=11, \\ \mathrm{probability} \mathrm{that} \mathrm{at} \mathrm{the} \mathrm{end} \mathrm{of} \mathrm{eleven} \mathrm{steps}, \\ \mathrm{he} \mathrm{is} \mathrm{just} \mathrm{one} \mathrm{step} \mathrm{away}=\mathrm{p}(\mathrm{x}=6)+\mathrm{p}(\mathrm{x}=5) \\ (\sin ce the number of successes =6 ,then number of failures =5 or \\ the number of successes =5, then number of failures =6) \\ ={ }^{11}{C}_6{\left(0.4\right)}^6{\left(0.6\right)}^4{+}^{11}{C}_5{\left(0.4\right)}^5{\left(0.6\right)}^5 \\ {=}^{11}{C}_5{\left(0.4\right)}^5{\left(0.6\right)}^5\left(0.4+0.6\right){=}^{11}{C}_5{\left(0.24\right)}^5 \end{gathered}$$