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# How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%?

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Toolbox:
• A random variable $X$ following binomial distribution with parameter n and p its probability of 'r' succeses $\rightarrow$ $P(X=r)=^nC_{r} p^{r} q^{n-r}$, where p is probability of success and $q=1-p$ and $r=0,1,\dots,n$
In any fair coin toss, P (getting a head) = P (getting a tail) $\rightarrow p = q = \large\frac{1}{2}$
We need to find $n$ such that the probability of getting at least one head is more than 90%
$P (X \geq 1) = 1 - P (X \lt 1) \gt 90$%
$\Rightarrow 1 - P (X = 0) \gt \large\frac{9}{10}$$\rightarrow P (X = 0) \lt 1 - \large\frac{9}{10}$$ \rightarrow P (X = 0) \lt\large \frac{1}{10}$
For a bionomial distribution, $P(X=0)=^nC_{0} \large\frac{1}{2}^{0} \frac{1}{2}^{n-0} $$=\large (\frac{1}{2})^{n} \Rightarrow \large( \frac{1}{2})^{n} \lt \frac{1}{10}$$ \rightarrow 2^n \gt 10$
Since $2^1 = 2, 2^3 = 8, 2^4 = 16$, the minimum value for $n$ that satifies the inequality is $n=4$, i.e, the coin should be tossed 4 or more times.
answered Jun 22, 2013
edited Feb 11, 2014