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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.

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    $\begingroup$ 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/…. $\endgroup$
    – whuber
    Feb 21, 2019 at 13:23

2 Answers 2

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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)
[1] 0.2401 0.4116 0.2646 0.0756 0.0081

Which confirms your calculation.

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The answer by Pere is right. If you want to get the result by hand, if we make n independent experiments with a success probability of p for each of them, the probability of getting exactly k successful (ie: "read-only") results is calculated as follows:

P(k) = p^k *(1-p)^(n-k) * n!/(k! * (n-k)! )

First factor p^k is the probability of getting k consecutive successes. Second factor (1-p)^(n-k) is the probability of getting (n-k) consecutive faillures. The final term n!/(k! * (n-k)! ) stands for the possible permutations of the results (it's there because we don't really care if successes and faillures are consecutive or not)

In conclusion, I guess you got it right!

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