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In order to get a head at least once with probability $\geq 0.9$, the minimum number of times a unbiased coin needs to be tossed is
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Verified Answer
The correct answer is:
4
We have, $P(H)=\frac{1}{2}, P(T)=\frac{1}{2}$
Now, $P(X \geq 1)=1-P(X=0$
$=1-{ }^{n} C_{o}(p)^{0}(q)^{n}=1-\left(\frac{1}{2}\right)^{n}$
$1-\frac{1}{2^{n}} \geq 0.9$
$\Rightarrow$
$0.1 \geq \frac{1}{2 n}$
$\Rightarrow$
$\frac{1}{10} \geq \frac{1}{2^{n}}$
$\Rightarrow$
$z^{n} \geq 10$
$\Rightarrow$
$n \geq 4$
Minimum number of tossed = 4
Now, $P(X \geq 1)=1-P(X=0$
$=1-{ }^{n} C_{o}(p)^{0}(q)^{n}=1-\left(\frac{1}{2}\right)^{n}$
$1-\frac{1}{2^{n}} \geq 0.9$
$\Rightarrow$
$0.1 \geq \frac{1}{2 n}$
$\Rightarrow$
$\frac{1}{10} \geq \frac{1}{2^{n}}$
$\Rightarrow$
$z^{n} \geq 10$
$\Rightarrow$
$n \geq 4$
Minimum number of tossed = 4
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