Explanation
$$\textbf{Step -1: Finding the probability of getting head in each toss.}$$
$$\text{Let a coin is tossed }n\text{ times.}$$
$$\text{The probability of getting head in one trial, }p=\dfrac{1}{2}.$$
$$\textbf{Step -2: Finding the number of times a coin is tossed.}$$
$$\text{Probability of getting at least one head}=P(X\geq1)$$
$$=1-P(X=0)$$
$$=1-^{n}C_{0}(1-p)^n$$
$$=1-\left(1-\dfrac{1}{2}\right)^n$$
$$=1-\dfrac{1}{2^n}\geq0.8$$ $$\text{(Given)}$$
$$\Rightarrow \dfrac{1}{2^n}\leq1-0.8$$
$$\Rightarrow \dfrac{1}{2^n}\leq0.2$$
$$\text{When }n=1,$$
$$\dfrac{1}{2^n}=\dfrac{1}{2}=0.5>0.2$$
$$\text{When }n=2,$$
$$\dfrac{1}{2^n}=\dfrac{1}{2^2}=0.25>0.2$$
$$\text{When }n=3,$$
$$\dfrac{1}{2^n}=\dfrac{1}{2^3}=0.125<0.2$$
$$\therefore n\text{ should be }3.$$
$$\textbf{Hence, correct option is D.}$$
$$ {\textbf{Step -1: Finding the Probability of success and failure}} $$
$$ {\text{Let a step forward be a success and a step backward be a failure}}{\text{.}} $$
$$ {\text{Then, the probability of success in one step}} $$
$$ \Rightarrow {\text{P = 0}}{\text{.4 = }}\dfrac{2}{5} $$
$$ {\text{The probability of failure in one step = Q = 0}}{\text{.6 = }}\dfrac{3}{5} $$
$$ {\text{In 11 steps he will be one step away from the staring point if the numbers of successes }}$$
$${\text{ and failure differ by 1}}{\text{.}} $$
$$ {\text{So, the number of successes = 6}} $$
$$ {\text{The number of failures = 5}} $$
$$ {\text{or the number of successes = 5,}} $$
$$ {\text{The number of failures = 6}} $$
$$ {\textbf{Step -2: Finding required probability}} $$
$$ \therefore {\text{The required probability = }}{}^{{\text{11}}}{{\text{C}}_{\text{6}}}{{\text{P}}^{\text{6}}}{{\text{Q}}^{\text{5}}} + {}^{{\text{11}}}{{\text{C}}_5}{{\text{P}}^5}{{\text{Q}}^6} $$
$$ = {}^{{\text{11}}}{{\text{C}}_{\text{6}}}{\left( {\dfrac{2}{5}} \right)^6}.{\left( {\dfrac{3}{5}} \right)^5} + {}^{{\text{11}}}{{\text{C}}_5}{\left( {\dfrac{2}{5}} \right)^5}.{\left( {\dfrac{3}{5}} \right)^6} $$
$$ = \dfrac{{11!}}{{6!.5!}}.{\left( {\dfrac{2}{5}} \right)^5}.{\left( {\dfrac{3}{5}} \right)^5}.\left\{ {\dfrac{2}{5} + \dfrac{3}{5}} \right\} $$
$$ = \dfrac{{11.10.9.8.7}}{{120}}.\dfrac{{{2^5}{{.3}^5}}}{{{5^{10}}}} $$
$$ = 462 \times {\left( {\dfrac{6}{{25}}} \right)^5} $$
$$ = 0.368 $$
$$ {\textbf{Hence, correct answer is option A}} $$
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