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 5 steps, he is one step away from the starting point.

Answer

P (step forward) $= \frac{2}{5},$ P (step backword)$= \frac{3}{5}$
He can remain a step away in either of the ways : 3 steps forward & 2 backwards 1 m or 2 steps forward & 3 backwards
$\therefore \text{required possibility} = ^5{\text{C}_{3}} \bigg(\frac{2}{5}\bigg)^{3}\bigg(\frac{3}{5}\bigg)^{2} + ^5{\text{C}}_{2}\bigg(\frac{2}{5}\bigg)^{2} \bigg(\frac{3}{5}\bigg)^{3}$
$= \frac{72}{125}$

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