Explanation
Step -1: Finding the probability of getting head in each toss.
Let a coin is tossed n times.
The probability of getting head in one trial, p=12.
Step -2: Finding the number of times a coin is tossed.
Probability of getting at least one head=P(X≥1)
=1−P(X=0)
=1−nC0(1−p)n
=1−(1−12)n
=1−12n≥0.8 (Given)
⇒12n≤1−0.8
⇒12n≤0.2
When n=1,
12n=12=0.5>0.2
When n=2,
12n=122=0.25>0.2
When n=3,
12n=123=0.125<0.2
∴n 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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