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 the starting point is
A). $${}^{11}C_{6}\left( 0.24\right)^{5} $$
B). $${}^{11}C_{6}\left( 0.4\right)^{6} \left( 0.6\right)^{5} $$
C). $${}^{11}C_{6}\left( 0.6\right)^{6} \left( 0.4\right)^{5} $$
D). None of these

Answer 1

Hint: In this question it is given that a man takes a step forward with probability 0.4 and backward with probability 0.6. We have to find the probability that at the end of eleven steps. So to find the solution we need to know that if we know the probability of success(p) and failure(q), then the probability of an event ‘p’ occurring exactly ‘r’ times is $$\mathrm{P} =\ ^{n} C_{r\ }p^{r}q^{n-r}$$.............(1)
Where, p+q=1,
 n = number of trials
 r = number of specific events you wish to obtain
 p = probability that the event will occur
 q = probability that the event will not occur

Complete step-by-step solution:
Since in this question a 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 and the probability of failure in one step(q) = 0.6.
Where p+q = 0.4+0.6 = 1.
Now let us denote a forward step by F and a backward step by B.
There are only two ways to take 11 steps and end up one step from the starting point:
Case1: F B F B F B F B F B F, or any permutation of this. The final position is one step ahead of the starting point.
Case 2: B F B F B F B F B F B, or any permutation of this. The final position is one step behind the starting point.
Therefore, from the above we can say that the number of successes is 6 and number of failures is 5 (from case 1) or the number of successes is 5 and number of failures is 6 (from case 2).
So for case 1, n =11(no. of trials), r = 6(no. of success), p= 0.4 and q=0.6
Therefore by formula (1) we can write,
$$\mathrm{P}_{1} =\ ^{11} C_{6\ }\left( 0.4\right)^{6} \left( 0.6\right)^{11-6} $$
   $$=\ ^{11} C_{6\ }\left( 0.4\right)^{6} \left( 0.6\right)^{5} $$...........(2)
For case 2, n =11(no. of trials), r = 5(no. of success), p= 0.4 and q=0.6
Therefore by formula (1) we can write,
$$\mathrm{P}_{2} =\ ^{11} C_{5\ }\left( 0.4\right)^{5} \left( 0.6\right)^{11-5} $$
   $$=\ ^{11} C_{5\ }\left( 0.4\right)^{5} \left( 0.6\right)^{6} $$..............(3)
Now since case 1 and 2 are independent upon each other, therefore by the fundamental principle of addition we can say that the probability that at the end of eleven steps, he is one step away from the starting point is,
$$\mathrm{P} =\mathrm{P}_{1} +\mathrm{P}_{2}$$
  $$=\ ^{11} C_{6}\left( 0.4\right)^{6} \left( 0.6\right)^{5} +\ ^{11} C_{5}\left( 0.4\right)^{5} \left( 0.6\right)^{6} $$
  $$=\ ^{11} C_{5}\left( 0.4\right)^{6} \left( 0.6\right)^{5} +\ ^{11} C_{5}\left( 0.4\right)^{5} \left( 0.6\right)^{6} $$ [$$\because {}^{n}C_{r}=\ ^{n} C_{n-r}$$]
  $$=\ ^{11} C_{5}\left( 0.4\right)^{5} \left( 0.6\right)^{5} \left( 0.4+0.6\right) $$
  $$=\ ^{11} C_{5}\left( 0.4\right)^{5} \left( 0.6\right)^{5} \left( 1\right) $$
  $$=\ ^{11} C_{5}\left( 0.4\right)^{5} \left( 0.6\right)^{5} $$
Which is our required probability.
So the correct option is option D.

Note: To solve this type of question you need to know about ‘Fundamental principle of addition’, which states that, if we have A ways of doing something and B ways of doing another thing and we can not do both at the same time, then there are A + B ways to choose one of the actions.