Minimum number of times a fair coin must be tossed so that the probability of getting at least one head is…

Let coin is tossed $n$ times.

The probability of getting a head in a single throw of a die is $\frac{1}{2}.$

By using binomial theorem, we know that the probability that an event $X$ appear $r$ times in $n$ trials is $P\left(X=r\right)={}^{n}C_r{p}^{n-r}{q}^r, q=1-p, r=0, 1, 2, ..., n.$

Now, $P($at least are head$)=1-P($no head$)$

$=1-{\left(\frac{1}{2}\right)}^n$

Using, the given condition, we get

$1-{\left(\frac{1}{2}\right)}^n\ge 0.99$

$\Rightarrow {\left(\frac{1}{2}\right)}^n\le 1-0.99$

$\Rightarrow {\left(\frac{1}{2}\right)}^n\le \frac{1}{100}$

As, $2^6=64$ and $2^7=128,$ hence, the minimum value of $n$ is $7.$