Multiplying probablity I am trying to help my kid do the following probability math. The language of the math baffles me. If the probability of the pictures tunring out is 1/5 then how can it become 3/4 (howeven numbers are taken) Any help in understanding the problem and a clue to solve it will be appreciated.
When Trilisa takes pictures, they turn out with probability $\frac{1}{5}$. She wants to take enough pictures so that the probability of at least one turning out is at least $\frac{3}{4}$. How few pictures can she take to accomplish this?
 A: When a picture turns out with probability $\frac 15$, it means they don't turn out with probability $\frac 45$.
Likewise, saying the probability that at least one turns out is $\frac 34$ is the same as saying that the probability that none turns out is $\frac 14$.
$(\frac 45)^n=\frac 14$ is the equation you have to solve. 
A: If I took a million pictures, it's really likely that one of them has turned out. I'll use a coins analogy instead: suppose I flip a coin which is heads with probability 1/5. I want to flip enough coins that the probability of at least one head is at least 3/4. Well, each time I flip a coin, I get more and more likely to get a head. If I flip ten coins, I need to have got ten tails in order to have not got at least one head, and it's actually not very likely that I get ten tails in ten coin flips (with p=1/5 of heads).
Clue to solving it: "at least one head" is precisely the complementary event of "no heads at all". What is the probability of no heads at all, in terms of the number of coin-tosses? ("Heads" -> "turn out", "coin-tosses" -> "pictures", if you like.)
A: The probability of no turning out is demanded to be at most $1-\frac34=\frac14$.
If $n$ pictures are taken then this probability equals $\left(1-\frac15\right)^n=\left(\frac45\right)^n$.
So to be solved is the inequality $\left(\frac45\right)^n\leq\frac14$.
The smallest $n\in\mathbb N$ that satisfies this is the answer to the question.
