Question

A drunkard man takes a step forward with probability 0.6 and takes a step backwards with probability 0.4. He takes 9 steps in all. Find the probability that he is just one step away from the initial point. Do you think drinking habits can ruin one’s family life?

Answer 1

Hint : Use the binomial distribution formula $q = 1 - p$, where $q$is the probability of failure, and $p$is the probability of success. The binomial distribution is the probability of success or failure outcome in an experiment or survey that is repeated multiple times.
A binomial experiment is applicable when the experiment consists of ‘n’ identical trials; each trial results in two outcomes, either success or failure, the probability of success remain the same from trial to trial, and n trials are independent.

Complete step-by-step answer :
Given he takes 9 steps in all
We need to find the probability when he is one step away from the initial position, and this can be backward or the forward
Let a step forward be success rate given for one step as $p = 0.4 = \dfrac{4}{{10}} = \dfrac{2}{5}$
And backward step be a failure given for one step as $q = 0.6 = \dfrac{6}{{10}} = \dfrac{3}{5}$
The man takes 9 steps in all which can be either 5 steps forward and 4 steps backwards or 4 steps forward and 5 steps backwards,
Hence the required probability will be $P = {}^9{C_5}{p^5}{q^4} + {}^9{C_4}{p^4}{q^5}$
By solving

$$P = {}^9{C_5}{p^5}{q^4} + {}^9{C_4}{p^4}{q^5} \\\ = {}^9{C_5}{\left( {\dfrac{2}{5}} \right)^5}{\left( {\dfrac{3}{5}} \right)^4} + {}^9{C_4}{\left( {\dfrac{2}{5}} \right)^4}{\left( {\dfrac{3}{6}} \right)^5} \\\ = \dfrac{{9!}}{{4!5!}}\left( {\dfrac{{2.2.2.2.2}}{{5.5.5.5.5}}} \right)\left( {\dfrac{{3.3.3.3}}{{5.5.5.5}}} \right) + \dfrac{{9!}}{{4!5!}}\left( {\dfrac{{2.2.2.2}}{{5.5.5.5}}} \right)\left( {\dfrac{{3.3.3.3.3}}{{5.5.5.5.3}}} \right) \\\ = \dfrac{{9!}}{{4!5!}}\left[ {\left( {\dfrac{{2.2.2.2.2}}{{5.5.5.5.5}}} \right)\left( {\dfrac{{3.3.3.3}}{{5.5.5.5}}} \right) + \left( {\dfrac{{2.2.2.2}}{{5.5.5.5}}} \right)\left( {\dfrac{{3.3.3.3.3}}{{5.5.5.5.3}}} \right)} \right] \\\$$

By further solving we get

$$P = \dfrac{{9!}}{{4!5!}}\left[ {\left( {\dfrac{{2.2.2.2.2}}{{5.5.5.5.5}}} \right)\left( {\dfrac{{3.3.3.3}}{{5.5.5.5}}} \right) + \left( {\dfrac{{2.2.2.2}}{{5.5.5.5}}} \right)\left( {\dfrac{{3.3.3.3.3}}{{5.5.5.5.3}}} \right)} \right] \\\ = \dfrac{{9.8.7.6}}{{4.3.2.1}}\left[ {\dfrac{{32}}{{3125}} \times \dfrac{{81}}{{635}} + \dfrac{{16}}{{635}} \times \dfrac{{243}}{{3125}}} \right] \\\ = 9.4.7\left[ {\dfrac{{2592 + 3888}}{{3125 \times 635}}} \right] \\\ = 252 \times \dfrac{{6480}}{{1984375}} \\\ = 0.8 \\\$$

Hence the probability he is one step away after taking 9 steps $= 0.8$

Note : Students are advised to be clear while selecting the value of the probability of success (p) and the probability of failure (q) and use them wisely as they vary for each question.