Question

A man takes a step forward with probability $0.4$ and backward with probability $0.6$. The probability that at the end of eleven steps he is one step away from starting point is

• A. $\frac{2^5{.3}^5}{5^{10}}$
• B. $462\times {\left(\frac{6}{25}\right)}^5$
• C. $231\times \frac{3^5}{5^{10}}$
• D. None of these

Answer 1

Answer: (B)

$462\times {\left(\frac{6}{25}\right)}^5$

As $0.4+0.6=1$, the man either takes a step forward or a step backward. Let a step forward be a success and a step backward be a failure.
Then, the probability of success in one step
$=p=0.4=\frac{2}{5}$

The probability of failure in one step $=q=0.6=\frac{3}{5}$.
In $11$ steps he will be one step away from the staring point if the numbers of successes and failure differ by $1$.

So, the number of successes $=6$
The number of failures $=5$
or the number of successes $=5$,
The number of failures $=6$
$\therefore$ the required probability ${=}^{11}{C}_6{p}^6{q}^5{+}^{11}{C}_5{p}^5{q}^6$

${=}^{11}{C}_6{\left(\frac{2}{5}\right)}^6.{\left(\frac{3}{5}\right)}^5{+}^{11}{C}_5{\left(\frac{2}{5}\right)}^5.{\left(\frac{3}{5}\right)}^6$

$=\frac{11!}{6!5!}.{\left(\frac{2}{5}\right)}^5.{\left(\frac{3}{5}\right)}^5.\{\frac{2}{5}+\frac{3}{5}\}$

$=\frac{\mathrm{11.10.9.8.7}}{120}.\frac{2^5{.3}^5}{5^{10}}=462\times {\left(\frac{6}{25}\right)}^5$.