What is the probability of getting given number of occurrences with known average outcome There are $92000$ random choices. Each one is selection of value $\{0, 1\}$ with given probabilities $\{0.95, 0.05\}$. So on average in series of $92000$ we will get $4600$ values of $1$.
What is the probability that in a sequence of $92000$ choices we will get less than $2300$ values of $1$? How should I approach this problem?
 A: 
You can estimate this easily by having compared this to a normal
distribution which this will closely mimic. Your exact distribution is
binomial with a mean of np=92000⋅0.05−4600 and a variance of
npq=92000⋅0.05⋅0.95, the standard deviation then being the square root
of the variance, about 66.1 here. You are then asking for the
probability of lying at or below almost 35 standard deviations below
the mean. That is incredibly unlikely, probability less than 10−250.
More careful tracking of values can get you more specific, I rounded
several places.

–  JMoravitz  4 mins ago
A: You can estimate this easily by having compared this to a normal distribution which this will closely mimic. Your exact distribution is binomial with a mean of $np=92000⋅0.05=4600$ and a variance of $npq=92000\cdot 0.05\cdot 0.95$, the standard deviation then being the square root of the variance, about $66.1$ here. You are then asking for the probability of lying at or below almost $35$ standard deviations below the mean. That is incredibly unlikely, probability less than $10^{-250}$. More careful tracking of values can get you more specific, I rounded several places.
See more on wikipedia or various learning websites like online.stat.psu.edu, probability.oer.math.uconn.edu or others
