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Q.
A man takes a step forward with probability $0.4$ and one step backward with probability $0.6$, then the probability that at the end of eleven steps he is one step away from the starting point is
A man will be 1 step away from its original position, if he takes 5 steps forward and 6 steps backward or 6 steps forward and 5 steps backward.
Hence, required probability
$={ }^{11} C_{5}(0.4)^{5}(0.6)^{6}+{ }^{11} C_{6}(0.4)^{6}(0.6)^{5} $
$={ }^{11} C_{5}(0.4)^{5}(0.6)^{5}(0.6+0.4) $
$={ }^{11} C_{5}(0.24)^{5} \cdot 1 $
$={ }^{11} C_{6}(0.24)^{5} \left[\because{ }^{n} C_{r}={ }^{n} C_{n-t}\right]$