Breakdowns and Bad Cartridges: Choosing Between Poisson and Binomial
A small digital-printing shop in Thrissur keeps two kinds of records. First, the number of times its main press breaks down on a working day; over a long period this has averaged 2.5 breakdowns per day, and breakdowns seem to occur independently at a steady rate. Second, the cartridges it buys arrive in lots of 15, and the supplier's long-run defective rate is 8 per cent. The owner wants to plan: he keeps one spare service engineer on days with three or more breakdowns, and he wants to know how often that will be needed. He also wants the chance that a lot of 15 cartridges contains at most one defective, because at two or more he can claim a replacement from the supplier. A friend suggests he use the Poisson distribution for both problems, since 'Poisson works for counts'. The owner is unsure and asks for the model to be chosen and justified for each record separately, with the probabilities worked out.
1. Model the daily breakdowns, justify the model, and find the probability that the spare engineer is needed on a given day.
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Model: daily breakdown count $X\sim$ Poisson with $\lambda=2.5$.
Justification: the record is a count of rare events in a fixed interval (one working day); breakdowns occur independently; the average rate is steady; two breakdowns cannot occur at the same instant. These are exactly the Poisson conditions - a fixed upper limit on the count is absent, unlike a binomial.
$$P(X=x)=e^{-2.5}\frac{2.5^x}{x!},\qquad e^{-2.5}=0.0821$$
$$P(X=0)=0.0821,\quad P(X=1)=0.0821(2.5)=0.2052,\quad P(X=2)=0.0821\left(\frac{6.25}{2}\right)=0.2565$$
$$P(X\le2)=0.5438$$
$$P(X\ge3)=1-0.5438=0.4562$$
So the spare engineer is needed on about 46% of working days - roughly every second day, a real rostering cost, not a rare contingency. Note also that for a Poisson variable the mean and variance both equal 2.5; the owner should check the recorded variance, because a variance far above 2.5 would signal clustering (breakdowns arriving in bad spells) and the model would need revisiting.
2. Model the number of defective cartridges in a lot, justify the model, and find the probability of at most one defective.
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Model: with $n=15$ cartridges and $p=0.08$, the count of defectives $Y\sim$ Binomial$(15,0.08)$, since there is a fixed number of trials, each cartridge is defective or not, trials are independent, and $p$ is constant.
$$P(Y=y)=\binom{15}{y}(0.08)^y(0.92)^{15-y}$$
$$P(Y=0)=(0.92)^{15}=0.2863$$ $$P(Y=1)=15(0.08)(0.92)^{14}=1.2(0.3112)=0.3734$$
$$P(Y\le1)=0.2863+0.3734=0.6597$$
So there is about a 66% chance of at most one defective, and therefore about a 34% chance of being able to claim a replacement ($Y\ge2$).
One caveat worth flagging: independence is the weak point. If the 15 cartridges come from a single production batch, a shared manufacturing fault makes defectives cluster, the binomial understates the chance of several defectives, and the claim rate would be higher than 34%.
3. Evaluate the friend's suggestion to use the Poisson model for both records.
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The suggestion is half right and would be wrong if applied blindly.
Breakdowns: Poisson is the natural model - unbounded count, no fixed number of trials, steady rate, independent occurrences.
Cartridges: binomial is the correct model, because there is a fixed $n=15$ and a constant success probability. A Poisson variable has no upper limit, so it could in principle assign probability to 16 or 20 defectives in a lot of 15 - impossible. Poisson can be used only as an approximation to the binomial when $n$ is large and $p$ is small with $np$ moderate.
Here $np=15(0.08)=1.2$, so try $\lambda=1.2$:
$$P(Y\le1)\approx e^{-1.2}(1+1.2)=0.3010(2.2)=0.6622$$
against the exact 0.6597 - an error of about 0.0025. The approximation is close but unnecessary when the exact binomial is easy to compute at $n=15$. The defensible rule to state: choose the model from how the random variable is generated, not from the fact that both are counts; use the approximation only when computation demands it and the conditions $n$ large, $p$ small hold.