Solving for Parasitic Worm Eggs in Moose Feces using Probability Distribution

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SUMMARY

The discussion revolves around calculating the number of slides a lab technician must prepare to achieve a 99% probability of detecting parasitic worm eggs in moose feces. The problem utilizes the binomial probability distribution, where the probability of finding eggs in a slide is 20% (p=0.2) and the probability of not finding eggs is 80% (q=0.8). The solution involves determining the total number of slides (n) needed to ensure that the probability of at least one slide showing eggs is 99%, which can be expressed as 1 minus the probability of zero slides containing eggs. The equation simplifies when setting the number of successful trials (h) to zero, allowing for the application of logarithmic functions to solve for n.

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I am a senior ecology undergraduate, this problem came up on the last assignment for my population ecology class. It's been a long time since I've done any probability, but I remember a bit and hope I'm on the right track.

Homework Statement



There is a species of parasitic worm that inhabits the intestines of the moose, the eggs of this parasite exit the moose in the feces for all practical purposes every moose who has the parasite has eggs in the feces. The eggs however are not very numerous and only 20% of all slides prepared from an infected fecal sample will contain some. How many slides would a lab tech have to prepare and to examine per specimen, so that if a specimen is positive, there is a 99% probability that at least one slide will show eggs present in the slide?

Homework Equations



The Attempt at a Solution



I'm thinking that the binomial probability distribution would be appropriate as each trial (slide) has only two outcomes (positive, eggs present, 20%, p=0.2 or negative, eggs not present, 80%, q=0.8). p+q=1. The total number of slides = n. Since the problem is asking for probability of "at least one slide", it would be the additive probabilities of one slide + two slides +. . . + n slides (0.99) which is equal to the alternative 1-(the probability that 0 slides out of n contain eggs(0.01)). If my thinking is correct so far . . . I would be able to use this equation

ml4.gif
=0.01

If I were to go with the 0 successful trials route the equation would equal 0.01, n=total number of trials (unknown in this situation), h=0, p=0.8.

From here I am stuck, I can't figure out how to solve for n.

I hope I'm on the right track.

Thanks
 
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When you set h=0, the expression will simplify quite a bit. Then take the log of each side to bring the n down from the exponent.
 

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