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

Answer 1

In such questions, use the concept of Probability of occurrence of event. Let us consider the event of man either takes a step forward or a step backward. Also assume the step forward be a success and a step backward be a failure.

Complete step by step answer:
According to the question, it is given that a man takes a step forward with probability 0.4 and backward with probability 0.6 respectively. So, we have
$\Rightarrow 0.4 + 0.6 = 1$
Then, we can say that the probability of success in one step:
$\Rightarrow p = 0.4 = \dfrac{2}{5}$
So, for the second case, the probability of failure in one step:
$\Rightarrow q = 0.6 = \dfrac{3}{5}$
Let us assume that 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
Therefore, the required probability =$^{11}{C_6}{p^6}{q^5}{ + ^{11}}{C_5}{p^5}{q^6}$
${ \Rightarrow ^{11}}{C_6}{\left( {\dfrac{2}{5}} \right)^5} \times {\left( {\dfrac{3}{5}} \right)^5}{ + ^{11}}{C_5}{\left( {\dfrac{2}{5}} \right)^5} \times {\left( {\dfrac{3}{5}} \right)^6}$
\Rightarrow \dfrac{{11!}}{{6!5!}}{\left( {\dfrac{2}{5}} \right)^5} \times {\left( {\dfrac{3}{5}} \right)^5} \times \left\\{ {\dfrac{2}{5} + \dfrac{3}{5}} \right\\}
$\Rightarrow \dfrac{{11 \times 10 \times 9 \times 8 \times 7}}{{120}} \times \dfrac{{{2^5} \times {3^5}}}{{{5^{10}}}}$
$\Rightarrow 462 \times {\left( {\dfrac{6}{{25}}} \right)^5}$

So, the correct answer is “Option B”.

Note: These questions are tricky but not tough. Use the method of probability on combinations or selections. To calculate the probability of a combination and selections, you will need to consider the number of favorable outcomes over the number of total outcomes. Combinations and selections are used to calculate events where order doesn't matter.