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 one step away from the starting point is

${^{11}C}_6 (0.24)^5$

The man will be one step away from the starting point if (i) either he is one step ahead or (ii) one step behind the starting point.

∴ The required probability = P(i) + P(ii)

The man will be one step ahead at the end of eleven steps if he moves six step forward and five steps backward.

The probability of this event is ${^{11}C}_6(0.4)^6 (0.6)^5$

The man will be one step behind at the end of eleven steps if he moves six steps backward and five steps forward.

The probability of this event is ${^{11}C}_6(0.6)^6 (0.4)^5$

Hence the required probability $={^{11}C}_6(0.4)^6 (0.6)^5 +{^{11}C}_6(0.6)^6 (0.4)^5={^{11}C}_6(0.4)^5 (0.6)^5 (0.4+0.6)= {^{11}C}_6(0.24)^5.$