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A coin is tossed $n$ times the probability of getting head at least once is greater than 0.8 . Then, the least value of such $n$ is :
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Verified Answer
The correct answer is:
3
The probability of getting head $=\frac{1}{2}$.
The probability of getting head atleast once in $n$ times.
$=\frac{1}{2}+\frac{1}{2^2}+\ldots+\frac{1}{2^n}=\frac{\frac{1}{2}\left(1-\left(\frac{1}{2}\right)^n\right)}{1-\frac{1}{2}}$
$=1-\left(\frac{1}{2}\right)^n$
Given that,
$1-\left(\frac{1}{2}\right)^n>0.8 \Rightarrow\left(\frac{1}{2}\right)^n < 0.2$
$\Rightarrow \quad 2^n>\frac{1}{0.2} \Rightarrow 2^n>5$
least value of $n$ for which $2^n>5$ is $n=3$.
The probability of getting head atleast once in $n$ times.
$=\frac{1}{2}+\frac{1}{2^2}+\ldots+\frac{1}{2^n}=\frac{\frac{1}{2}\left(1-\left(\frac{1}{2}\right)^n\right)}{1-\frac{1}{2}}$
$=1-\left(\frac{1}{2}\right)^n$
Given that,
$1-\left(\frac{1}{2}\right)^n>0.8 \Rightarrow\left(\frac{1}{2}\right)^n < 0.2$
$\Rightarrow \quad 2^n>\frac{1}{0.2} \Rightarrow 2^n>5$
least value of $n$ for which $2^n>5$ is $n=3$.
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