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). \[{}^{11}{{C}_{5}}\cdot {{\left( 0.48 \right)}^{5}}\]
(b). \[{}^{11}{{C}_{6}}\cdot {{\left( 0.24 \right)}^{5}}\]
(c). \[{}^{11}{{C}_{5}}\cdot {{\left( 0.12 \right)}^{5}}\]
(d). \[{}^{11}{{C}_{6}}\cdot {{\left( 0.72 \right)}^{6}}\]

Answer 1

Hint: - As we can see that 0.4+0.6=1, hence, the man either moves forward or backwards. Now, in this question, there will be two cases as it is mentioned that the man has to be a step away from the starting point. Hence, the one step can be either backwards or forwards. The formula for choosing r things from n given things is as follows
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]


Complete step-by-step solution -
As mentioned in the question, we have to find the probability of the man to be one step away from the starting point.
Now, for case 1, we will consider the final position of that man to be one step forward.
Now, we will choose 6 steps out of the 11 to which we will assign a forward step as follows
\[={}^{11}{{C}_{6}}\]
Hence, the remaining 5 steps would be backward steps. Therefore, we can write as follows
\[={}^{11}{{C}_{6}}\cdot {{\left( 0.4 \right)}^{6}}\cdot {{\left( 0.6 \right)}^{5}}\]
Now, for case 2, we will consider the final position of that man to be one step backward.
Now, we will choose 6 steps out of the 11 to which we will assign a backward step as follows
\[={}^{11}{{C}_{6}}\]
Hence, the remaining 5 steps would be forward steps. Therefore, we can write as follows
\[={}^{11}{{C}_{6}}\cdot {{\left( 0.6 \right)}^{6}}\cdot {{\left( 0.4 \right)}^{5}}\]
Now, the total probability that is required is calculated as follows
     \[\begin{align}
  & ={}^{11}{{C}_{6}}\cdot {{\left( 0.6 \right)}^{6}}\cdot {{\left( 0.4 \right)}^{5}}+{}^{11}{{C}_{6}}\cdot {{\left( 0.4 \right)}^{6}}\cdot {{\left( 0.6 \right)}^{5}} \\
 & ={}^{11}{{C}_{6}}\cdot {{\left( 0.24 \right)}^{5}}\cdot \left( 0.4+0.6 \right) \\
 & ={}^{11}{{C}_{6}}\cdot {{\left( 0.24 \right)}^{5}} \\
\end{align}\]
Hence, this is the final answer to the question.

Note: -The students can make an error if they are not able to imagine the two cases separately as without this fact one could not get to the correct answer as it is the most important step indeed to solve the question.