# Checking my combinatorics

I am modelling a flash memory system where 0.3 requests are writes (and so take 100 cycles to complete) and 0.7 are read-only (and so take 50).

The system can handle 4 requests at once (the proportions are based on observed values but the timings and the 4-at-once are arbitrary).

By my calculation, and assuming the system is "fully loaded" this means 0.2401 probability of 4 read-only, 0.4116 of 1 write and 3 read-only, 0.2646 of 2 writes and 2 read-only, 0.0756 of 3 writes and 1 read-only and 0.0081 of 4 writes.

That all, happily, adds to 1 - but is it correct? My combinatorics is very rusty and online calculators don't allow for proportions.

• You might be amused by the coefficients in the polynomial $(3 + 7t)^4.$ if the connection with your question seems obscure, much more can be found on our site at stats.stackexchange.com/….
– whuber
Feb 21, 2019 at 13:23

Assuming that you have 4 request, that each of them have a probability 0.3 of being a write request and that the events of each of them being a write request are independent, the number of write requests follow a binomial distribution with parameters n=4 and p=0.3.

The probabilities of 0 to 4 write requests are:

> dbinom(0:4, 4, .3)
 0.2401 0.4116 0.2646 0.0756 0.0081