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)

${}^{11}{C}_6\times (0.24{)}^5$

Let a step forward be a success and a step backward be a failure.

Then the probability of success in one step is: $p=0.4=\frac{4}{10}=\frac{2}{5}$

And the probability of failure in one step is:$q=0.6=\frac{6}{10}=\frac{3}{5}$

In eleven steps he will be one step away from the starting point if the number of successes and failures differs by $1$.

Therefore, the number of successes is $6$ and number of failures is $5$ or the number of successes is $5$ and number of failures is $6$.

Thus, the required probability is

$={{}^{11}C}_6{p}^6{q}^5+{{}^{11}C}_5{p}^5{q}^6={{}^{11}C}_6{\left(\frac{2}{5}\right)}^6{\left(\frac{3}{5}\right)}^5+{{}^{11}C}_5{\left(\frac{2}{5}\right)}^5{\left(\frac{3}{5}\right)}^6={{}^{11}C}_6{\left(\frac{2}{5}\right)}^5{\left(\frac{3}{5}\right)}^5[\frac{3}{5}+\frac{2}{5}]$ $={{}^{11}C}_6{\left(\frac{6}{25}\right)}^5\left(1\right)={{}^{11}C}_6\times {\left(0.24\right)}^5$

Hence, the probability that at the end of eleven steps he is one step away from the starting point is ${{}^{11}C}_6\times {\left(0.24\right)}^5$.