How can I estimate the number of ads a snack bar owner needs to use along a road to achieve 99/100 of driver's attention I snack bar owner performed a research and noticed that with one ad on the road he achieved 1/2 of driver's attention (the ad was seen). He was unsatisfied he wanted a value above 99/100  how many more ads he must use at minimum?
a) 99
b) 51
c) 50
d) 6
e) 1

I don't even now how to solve it. I promese I tried but probability has no middleground or you know how to approach it or you just stand still. How can I approach a problem like this?
I corrected the question it is not how many but how many more ads he must use.
 A: What's given can be rephrased as "Each ad on the road gains $1/2$ of the drives' attention" now it might be tempting to say you need only $2$ ads to gain all the driver attention but that would be wrong, imagine you have $100$ drivers, if the first ad gains half the attention then you have the attention of $50$ drivers and there is another $50$ remaining, add another ad and you will have the attention of half of those remaining so $25$, in total you will have the attention of $75$ drivers with $25$ drivers remaining, etc...
So basically you have to find $k$ such that $\sum_{n=1}^{k} (\frac{1}{2})^{n} >0.99$
Since you are given options then the easiest thing to do would be to plug them in for $k$, we have $\sum_{n=1}^{6} (\frac{1}{2})^{n}=0.9843...$ which is still below our target.
We have $\sum_{n=1}^{50} (\frac{1}{2})^{n}=1$ which is above our target, so the answer is $50$.
Note that to find the sum you have two options, either put the series in a calculator or since it is a geometric series you can do the following:
$S_n=a_1(\frac{(1−r^n)}{1−r}), r\neq1$
Where $n$ is the number of terms, $a_1$ is the first term and $r$ is the common ratio
so for $\sum_{n=1}^{6} (\frac{1}{2})^{n}$ we have $n=6, r=\frac{1}{2}, a_1=\frac{1}{2}$ so our sum would be
$\frac{\frac{1}{2}(1-(\frac{1}{2})^6)}{1-\frac{1}{2}}=\frac{\frac{1}{2}(\frac{63}{64})}{\frac{1}{2}}=\frac{63}{64}=0.9843...$
NOTE: this is the minimum without taking into account there is an ad already placed, so if we have no ads at all, BETWEEN THESE OPTIONS we would need to choose $50$, however, if the question actually asks for how many ads other than the one already placed we need then we would go for option of $6$ ads as $7$ ads are enough to get our desired number, we have: $\sum_{n=1}^{7} (\frac{1}{2})^{n}=0.9921875$
A: It might be assumed that each ad will get $\frac{1}{2}$ of drivers' attention as well, which means missing the other $0.5$. Also, assume that the events of any driver see one ad and another ad are irrelevant. Then the ratio of drivers haven't seen an ad after $n$ ads is $0.5^n$. The goal is to find the minimal $n$ such that $$0.5^n<0.01$$
$$n>\frac{lg(0.01)}{lg(0.5)}\approx6.64$$
Or check that $\left(\frac{1}{2}\right)^6 = \frac{1}{64}, \left(\frac{1}{2}\right)^7 = \frac{1}{128}, 7$ is enough.
The minimal number of ads should be 7.
A: After placing $n$ ads on the road, $\frac1{2^n}$ of drivers have not seen at least one ad, assuming that drivers see each ad with equal probability. This falls below the target value of $\frac1{100}$ at seven ads, so you should put $6$ as the answer.
