What is the number of times shooter needs to fire?

Question

A shooter firing at a target has 10% chance of hitting the target in one shot. The number of times he must fire at the target to have above 50% chance of hitting the target is:
How do I solve this problem?
P.S.:I was thinking to apply binomial distribution to this problem,but I have no idea how to apply it here.

Answer 1

Let $n$ be the number of shots that need to be fired. The probability of hitting the target in a shot is $10\%$; so the probability of missing the target in a shot is $(1-10\%)$. The probability of not missing the target in all $n$ shots is $1 - (1-10\%)^n$, which we require to exceed $50\%$, i.e., $$1 - (1-10\%)^n > 50\% \Rightarrow n \geq \left\lceil \frac{\log 0.5}{\log0.9}\right\rceil =7.$$

Answer 2

We seek the minimum count of shots, $k$, where the probability for hitting the target within them is at least $50\%$.   That is also where the probability for missing the target within $k$ shots being at most $50\%$.

Since the probability for missing any particular shot is $0.9$, then the probability for missing the target with $k$ shots is: $P(k) = [\text{some function of }k]$, but what?

Having determined that, find $\min\{k\in\Bbb N^+: P(k)\leq 0.50\}$ .

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PS: The count of (iid Bernoulli) trials until a success follows a Geometric$_1$ Distribution.