modelling using Binomial p.d.f I am trying to get expected number of days for one to get a loss of more than 5 Yuan, if it costs 2 Yuan to use a coffee vendor machine that has a success rate of 99% per vending. I try the machine twice per day. 
What is the probability of success on two occasions , is it 0.99x0.99?
Thanks
 A: I would model the problem using the negative binomial, or more simply the geometric distribution. We want the expected time until the third failure (third because we are losing $2$ Yuan each time).
The expected number of trials until the first loss is $\frac{1}{0.01}$. So the expected number until the third loss is $\frac{3}{0.01}$, that is, $300$. That makes the expected number of days $150$.
A: Expected Value, $E(x)$ for Binomial Distribution is np, where n is number of trials and p is probability of "success". 
*note that "success" in this question is the probability of failure of the machine where you lose 2 yuan.
$E(x)=np$
In this case, p is probability of failure (machine not working). $p=0.01$
$n$ is what we need to find out. n represents the number of failed attempts where you lose your 2 yuan.
If E(x) = 5 yuan, $5=(0.01)*(n)$
Thus, n equals 500 trials, which equals 250 days since you try 2 times per day. As the question asks for loss of more than 5 yuan, more than 250 days should be expected before a loss of more than 5 yuan.  
