Question
A laboratory wants to estimate how many seed packets are contaminated. It tests 40 disjoint pools of four packets each. A pool is reported positive if at least one of its four packets is contaminated; the assay has no false positives or false negatives. Each packet is contaminated independently with the same unknown probability \(p\). Exactly 15 pools test positive, but the laboratory does not know which packets within them were contaminated.
Find the maximum-likelihood estimate of \(p\). Explain why \(\frac{15}{40}\) is not the estimated packet contamination rate. What estimates result if none of the pools, or all of them, test positive?
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Hint 1
A pool is negative only if all four members are clean; translate p into a positive-pool probability \(\theta\).
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Solution
A packet is clean with probability \(1-p\). Because packet contamination events are independent and pools do not overlap, a four-packet pool is negative with probability \((1-p)^{4}\) and positive with probability \(\theta(p)=1-(1-p)^{4}\). The number Y of positive pools is therefore \(\operatorname{Binomial}(40,\theta(p))\). For the observed \(Y=15\), its likelihood, up to the binomial coefficient, is \(L(p)=[1-(1-p)^{4}]^{15}[(1-p)^{4}]^{25}\).
The binomial likelihood in θ is maximised at \(\hat\theta=\frac{15}{40}=\frac{3}{8}\). The function \(\theta(p)\) is strictly increasing on \(0\le p\le1\), so the corresponding MLE of p solves \((1-p)^{4}=1-\frac{3}{8}=\frac{5}{8}\). Therefore \(\hat p=1-\left(\frac{5}{8}\right)^{1/4}\approx0.1108603\). The fraction \(\frac{15}{40}\) estimates the probability a four-packet union test is positive, not the probability an individual packet is positive; positive pools can contain one, two, three or four contaminated packets.
If \(Y=0\), the likelihood is \((1-p)^{160}\), which is maximised at \(p=0\). If \(Y=40\), the likelihood is \([1-(1-p)^{4}]^{40}\), maximised at \(p=1\). These are legitimate boundary maximum-likelihood estimates, not evidence that the underlying rate has been proved exactly zero or one.
Assumptions
The 160 packets are independent Bernoulli(p) items and are divided into 40 nonoverlapping four-packet pools.
The assay is perfect and detects the presence of at least one contaminated member, never the count.
All packets share the same unknown \(p\) in \([0,1]\).
Final Answer
\(\hat p=1-\left(\frac{25}{40}\right)^{1/4}=1-\left(\frac{5}{8}\right)^{1/4}\approx0.11086\) (11.09%). With 0 positive pools \(\hat p=0\); with all 40 positive \(\hat p=1\).