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)

$^{11}C_6\times(0.24)^5$

Answer 2

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 $={ }^{11} C_{5}(0.4)^{5}(0.6)^{6}+{ }^{11} C_{6}(0.4)^{6}(0.6)^{5}$ $={ }^{11} C_{5}(0.4)^{5}(0.6)^{5}(0.6+0.4)$ $={ }^{11} C_{5}(0.24)^{5} \cdot 1$ $={ }^{11} C_{6}(0.24)^{5} \quad\left[\because{ }^{n} C_{r}={ }^{n} C_{n-t}\right]$