Question

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. ${}^{11}{C}_5\times (0.48{)}^5$
• B. ${}^{11}{C}_6\times (0.24{)}^5$
• C. ${}^{11}{C}_5\times (0.12{)}^5$
• D. ${}^{11}{C}_6\times (0.72{)}^6$

Answer 1

Answer: (B)

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
$={}^{11}{C}_5(0.4{)}^5(0.6{)}^6+{}^{11}{C}_6(0.4{)}^6(0.6{)}^5$
$={}^{11}{C}_5(0.4{)}^5(0.6{)}^5(0.6+0.4)$
$={}^{11}{C}_5(0.24{)}^5\cdot 1$
$={}^{11}{C}_6(0.24{)}^5\left[\because {}^n{C}_r={}^n{C}_{n-t}\right]$