Probabilitical Steps..

Since the man is one step away from starting point it means that either

(i) man has taken 6 steps forward and 5 steps backward.

(ii) man has taken 5 steps forward and 6 steps backward.

Taking movement of 1 step forward as success and movement of 1 step backward as failure.

p = Probability of success = 0.4

q = Probability of failure = 0.6

Applying Binomial Theoram, $Required \ probability = P(X=6 \ to \ X = 5)$ $= P(X = 6) + P(X = 5)$ $= {11 \choose 6}p^6q^5 + {11 \choose 5}p^5q^6$ $= {11 \choose 5}(p^6q^5 + p^5q^6)$ $= \frac{11.10.9.8.7}{1.2.3.4.5}[(0.4)^6(0.6)^5 + (0.4)^5(0.6)^6]$ $= \boxed{0.37}$

The man has to finish one step away from starting point so he has either taken:

5 steps forward and 6 steps back

Or 6 steps forward and 5 steps back

For each arrangement of 5 steps forward and 6 steps back, the probability of that happening is:

$(0.4)^5$ x $(0.6)^6$

And there are 11C5 ways to place the 5 forward steps within the 11 steps so the total probability for this case is:

$(0.4)^5$ x $(0.6)^6$ x (11C5)

And for each arrangement of 6 steps forward and 5 steps back the probability is:

$(0.4)^6$ x $(0.6)^5$

And there are 11C6 ways to place the 6 forward steps (which is equal to 11C5 ways to place the backward steps) so the probability of this case is:

$(0.4)^6$ x $(0.6)^5$ x (11C6)

So the total probability is the two cases added together which equals:

$((0.4)^5$ x $(0.6)^6$ x (11C5)) + ( $(0.4)^6$ x $(0.6)^5$ x (11C6)) = $\boxed{0.37}$ to 2dp.