Question

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

• A. $=462\times (0.4{)}^6\times (0.6{)}^4$
• B. $=462\times (0.24{)}^{10}$
• C. $=462\times (0.4{)}^4\times (0.6{)}^6$
• D. $=462\times (0.24{)}^5$

Answer 1

The correct option is D

$=462\times (0.24{)}^5$

The man will be one step away from the starting point if
(i) either he is one step ahead or
(ii) one step behing the starting point.

Now if at the end of eleven steps the man is one step ahead of the starting point, he must take six steps forward and five steps backwards.
The probability of this event
$={{}^{11}C}_6\times (0.4{)}^6\times (0.6{)}^5=462\times (0.4{)}^6\times (0.6{)}^5$
Again if at the end of eleven steps the man is one step behind the starting point, then out of 11 steps he must have taken six steps backward and five steps forward.
The probability of this event
$={{}^{11}C}_6\times (0.6{)}^6\times (0.4{)}^5=462\times (0.6{)}^6\times (0.4{)}^5$

Since the events (i) and (ii) are mutually exclusive, the probability that one of these event happens
$=462\times (0.4{)}^6\times (0.6{)}^5+462\times (0.6{)}^6\times (0.4{)}^5=462\times (0.4{)}^5\times (0.6{)}^5[0.4+0.6]=462\times (0.4\times 0.6{)}^5\times 1=462\times (0.24{)}^5$