Probabilities involving "at least one" success Isabella built a time travel machine, but she can't control the destination of her trip. Each time she uses the machine she has a $.25$ probability of traveling to a time before she was born. During the first year of testing, Isabella uses her machine $5$ times.
Assuming that each trip is equally likely to travel before Isabella was born, what is the probability that at least one trip will travel before Isabella was born?
Round your answer to the nearest hundredth.
I get the answer $1-.75^4$
However, the correct answer is $1-.75^5$. How is this the answer? If finding the complement wouldn't we account for one of the times being before she is born?
 A: We do not need to account for probability that she was born, it is known to us.  
You want at least $1$ case, so why not calculate none, or $0$ case, and subtract it from total number of cases?  
Or simply, $$P(at least\hspace{0.2cm} 1)=1 - P(none)  $$
$P(none)$ means calculate probability that she always goes in future.
$P(future)=0.75$ for $1$ trip.
So, for $5$ trips, it will be $0.75^5$.
Subtract this from $1$:   
You get $$1-0.75^5$$

If this confuses you, calculate discretely for cases: exactly once, exactly twice,thrice, four times and five times.
$$P(exactly \hspace{0.2cm} once)=(0.25)\cdot (0.75)^4$$
$$P(exactly \hspace{0.2cm} twice)=(0.25)^2\cdot (0.75)^3$$
$$P(exactly \hspace{0.2cm} thrice)=(0.25)^3\cdot (0.75)^2$$
$$P(exactly \hspace{0.2cm} 4 \hspace{0.2cm} times)=(0.25)^4\cdot (0.75)^1$$
$$P(exactly \hspace{0.2cm} 5 \hspace{0.2cm} times)=(0.25)^5\cdot (0.75)^0$$
Just add all probabilities: $$(0.25)^1\cdot (0.75)^4+(0.25)^2\cdot (0.75)^3+(0.25)^3\cdot (0.75)^2+(0.25)^4\cdot (0.75)^1+(0.25)^5\cdot (0.75)^0$$
Verify if you get same result!
