Probability theory basis-Poisson distribution calculation of approximate Probability

 
To ensure the normal operation of the equipment, a certain number of equipment maintenance personnel are required. There are 300 similar equipment and each equipment works independently. The failure probability at any time is 0.01, if one person repairs a device fault, how many repair personnel should be assigned at least to ensure that the probability of timely repair after the device failure is not less than 0.99?
 

When N is large and P is small, this formula can be used to calculate the distribution of two similar items. The calculation amount is less.

Of course, for this question, exact calculation should use two distributions, but the calculation of C (n, k) is too large and will overflow.

The followingProgramIt is concluded that at least 8 persons are required.

# Include <iostream> # include <math. h> using namespace STD; int M [101] = {300}; int main () {int n = 0.01; Double P =; double r = N * P; // calculate the factorial. The prediction result is not too large. You can calculate it to 50 for (INT I = 2; I <= 50; I ++) M [I] = m [I-1] * I; double KE = exp (-R); double ans = 0; // calculate for (INT I = 0 Using Poisson distribution formula; I <300; I ++) {ans + = ke * POW (R, I)/m [I]; If (ANS> = 0.99) {cout <I <Endl; break ;}} return 0 ;}