Bernoulli trial given probability of success - missile hitting a target problem A missile has a $25$% chance of hitting a target. How many missiles  must be fired to ensure that there is a greater than $90$% chance that the target will be hit at least once? 
Been using the binomial distribution formula, but can't seem to get the answer?
 A: Let $X$ be the number of times the target is hit, $n$ be the number of missiles fired, and $p=0.25$ be the probability that a missile hits the target. Then $X\sim \text{Binom}(n,p)$. 
You wish to find the smallest $n$ so that $P(X\geq 1) \geq 0.9$. So we must calculate $P(X \geq 1)$. The easiest way to do this is to write it as $P(X \geq 1) = 1-P(X=0)$. Then the problem becomes finding the smallest $n$ so that $P(X=0) \leq 0.1$. Can you proceed from there?
A: Alternative basic approach. If there is a greater than $\frac{9}{10}$ chance of hitting the target at least once, there must be a less than $\frac{1}{10}$ chance of never hitting the target.  Then:


*

*A single missile has a $\frac34$ chance of missing the target.

*Two missiles have a $\frac34 \times \frac34 = \frac{9}{16}$ chance of missing the target.

*Three missiles have a $\left(\frac34\right)^3 = \frac{27}{64}$ chance of missing the target.


When does this probability drop below $\frac{1}{10}$?  Use logs to finish this answer off.
ETA: We assume the missiles' chances are independent.  In real life, this isn't necessarily a warranted assumption.
A: You want $P(\text{at least one success}) \geq 0.9$
This is the same as $1-P(\text{no successes}) \geq 0.9$
Then you can find that $P(\text{no successes}) \leq 0.1$
where the sign flip because we divide both sides by $-1$.
For an unknown number of trials $n$, we have that
$P(\text{no successes}) = .75^n$.
Then 
$$\begin{align*}
log(.75^n) \leq log(0.1)
& \Rightarrow nlog(.75) \leq log(0.1) \\\\
&\Rightarrow n \geq \frac{log(0.1)}{log(0.75)} \text{ (sign flips because } log(.75) \lt 0\\\\
&\Rightarrow n \geq 8.004
\end{align*}$$
But we cannot have $8.004$ trials which means it must be $9$.
Checking in R:
1-dbinom(0, 9, .25)

0.9249153

So this probability indeed exceeds $0.9$.
Note that we come really close to having a $90$% chance of hitting the target at least once if we fire $8$ missiles. 
1-dbinom(0, 8, .25)

0.8998871

