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.
$$\frac{{{2^5}{{.3}^5}}}{{{5^{10}}}}$$
B.
$$462 \times {\left( {\frac{6}{{25}}} \right)^5}$$
C.
$$231 \times \frac{{{3^5}}}{{{5^{10}}}}$$
D.
none of these
Answer :
$$462 \times {\left( {\frac{6}{{25}}} \right)^5}$$
Solution :
As $$0.4 + 0.6 = 1,$$ the 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 $$ = p = 0.4 = \frac{2}{5}$$
The probability of failure in one step $$ = q = 0.6 = \frac{3}{5}$$
In $$11$$ steps he will be one step away from the starting point if the numbers of successes and failures differ by $$1.$$
So, the number of successes $$ = 6.$$ The number of failures $$ = 5.$$
or the number of successes $$ = 5.$$ The number of failures $$= 6.$$
$$\therefore $$ The required probability
$$\eqalign{
& = {}^{11}{C_6}{p^6}{q^5} + {}^{11}{C_5}{p^5}{q^6} \cr
& = {}^{11}{C_6}{\left( {\frac{2}{5}} \right)^6}.{\left( {\frac{3}{5}} \right)^5} + {}^{11}{C_5}{\left( {\frac{2}{5}} \right)^5}.{\left( {\frac{3}{5}} \right)^6} \cr
& = 462 \times {\left( {\frac{6}{{25}}} \right)^5} \cr} $$