12
24
𝐿𝑖𝑘𝑒𝑙𝑖ℎ𝑜𝑜𝑑 = ∑ [( ) 0.117𝑖 (1 − 0.117)24−𝑖 ] 𝑓(𝑖, 𝑛, 𝑧)
𝑖
𝑖=0
We calculated the likelihood function starting with the probability of i number of
trials scoring positive because of the non-spiked oak water. This is a binomial
distribution with 24 total trials, and the probability of a single trial scoring positive
because of non-spiked oak water as 0.117, which was empirically measured. As we
used simulations to calculate the likelihood, we halted at i=12 trials scored positive
because of the non-spiked oak water, as more positive trials should occur with
probability < 10-6.
𝑓(𝑖, 𝑛, 𝑧) =
𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑛
24 − 𝑖
=(
) 𝑔(𝑧, 𝜆)𝑛−𝑖 (1 − 𝑔(𝑧, 𝜆))24−𝑛
𝑡𝑜𝑡𝑎𝑙 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑡𝑟𝑖𝑎𝑙𝑠
𝑛−𝑖
We calculated the probability of n total positive trials (as n – i positive trials caused
by spiked oak water) using a binomial distribution with 24 total trials and g(z, 𝜆) as
the probability of a positive trial.
10
𝜆𝑘 𝑒 −𝜆
𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑎 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑡𝑟𝑖𝑎𝑙
(1 − (1 − 𝑧)𝑘 )
𝑔(𝑧, 𝜆) =
= ∑
𝑐𝑎𝑢𝑠𝑒𝑑 𝑏𝑦 𝑠𝑝𝑖𝑘𝑒𝑑 𝑜𝑎𝑘 𝑤𝑎𝑡𝑒𝑟
𝑘!
𝑘=0
We calculated the probability of a positive trial using a Poisson distribution to
examine the number of spiked cells across wells. The average number of cells per
well, 𝜆, was empirically measured for each yeast strain. z is the probability of
detecting a single cell, but this probability is applied for each cell in each well. We
halted at k=10 cells per well as more cells should occur with probability < 10-6.
We simulated the model across values of z ranging from 0 to 2 by 0.001 increments
using the n and 𝜆 values derived from each yeast strain.