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 _____________

Answer 1

It is given in the question that the probability of taking a step forward and a step backward. We have to find the probability that at the end of eleven steps, he is one step away from the starting point. For that we will take the given process as success and failure, from the given eleven steps we will consider two possibilities and find the probabilities of two possibilities and sum them up to find the required probability.

Formula used: ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$.

Complete step-by-step answer:
The probability of event ${\bf{P}}\left( {\bf{E}} \right) + {\bf{P}}\left( {{\bf{E}}'} \right) = {\bf{1}}$ where $E$ is event and $E'$ represents that the event will not occur.
Complete step by step answer:
Here, it is given that 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
$p = 0.4 = \dfrac{4}{{10}} = \dfrac{2}{5}$
The probability of failure in one step $q = 0.6 = \dfrac{6}{{10}} = \dfrac{3}{5}$
In 11 steps he will be one step away from the starting point if the numbers of successes and failures 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
The general formula of probability distribution is given as ${}^nC_r p^n q^r$.
Where n is number of successes,r is number of failures, p is probability of success and q is probability of failure.
∴ The required probability $= {}^{11}{C_6}{p^6}{q^5} + {}^{11}{C_5}{p^5}{q^6}$
$= {}^{11}{C_6}{\left( {\dfrac{2}{5}} \right)^6} \cdot {\left( {\dfrac{3}{5}} \right)^5} + {}^{11}{C_5}{\left( {\dfrac{2}{5}} \right)^5} \cdot {\left( {\dfrac{3}{5}} \right)^6}$
Simplifying the equation,
= \dfrac{{11!}}{{6!5!}} \times {\left( {\dfrac{2}{5}} \right)^5} \times {\left( {\dfrac{3}{5}} \right)^5} \times \left\\{ {\dfrac{2}{5} + \dfrac{3}{5}} \right\\}
$= \dfrac{{11 \times 10 \times 9 \times 8 \times 7}}{{120}}\left( {\dfrac{{{2^5} \times {3^5}}}{{{5^{10}}}}} \right)\left( {\dfrac{5}{5}} \right)$
By solving the above we get,
$= 462 \times {\left( {\dfrac{6}{{25}}} \right)^5}$
Hence, the probability that at the end of eleven steps he is one step away from the starting point is $462 \times {\left( {\dfrac{6}{{25}}} \right)^5}$.

Note: The probability of an event lies between 0 and 1, where, roughly speaking, 0 indicates impossibility of the event and 1 indicates certainty. The higher the probability of an event, the more likely it is that the event will occur.Students should remember general probability distribution and formula to solve these types of questions.