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

${^{11}C}_5(0.4)^5(0.6)^5$

Let p denote the probability that the man takes a step forward. Then, p = 0.4

$∴ q=1-p=1-0.4 = 0.6$

Let X denote the number of steps taken in the forward direction. Since the steps are independent of each other, therefore X is a binomial variate with parameters n = 11 and p = 0.4 such that

$P(X=r) ={^{11}C}_r, (0.4)^r (0.6)^{11-r} = 0, 1, 2,..., 11$ ...(i)

Since the man is one step away from the initial point, he is either one step forward or one step backward from the initial point at the end of eleven steps. If he is one step forward, then he must have taken six steps forward and five steps backward and if he is one step backward, then he must have taken five steps forward and six steps backward. Thus, either X = 6 or X=5.

∴ Required probability = $P [(X = 5) or (X = 6)]$

$= P(X=5) + P(X = 6)$

$= {^{11}C}_5 (0.4)^5 (0.6)^{11-5}+{^{11}C}_6 (0.4)^6 (0.6)^{11-6}$  [Using (i)]

$={^{11}C}_5 (0.4)^5 (0.6)^5 [0.6 +0.4]$    $[∵ {^{11}C}_5= {^{11}C}_6]$

$={^{11}C}_5 (0.4)^5 (0.6)^5$