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A protein that has the objective of assisting pig growth is given to pigs in their lunch.
A researcher observes that few pigs may have developed tumours as a result of their eating which comprises the protein.
The researcher who added the protein discovers that for a random pig that gets tumours - there is a probability ranging from p=$frac110$, when a pig has a standard lunch
to p=$frac12$ when given the lunch that has the protein present.
The researchers' investigation has 100 pigs, where the number of pigs that are given the lunch with the protein present is unknown (I refer to this as n).

A second researcher randomly chooses a pig from the 100 pig cohort. If the pig has tumours, find the probability that the pig had been given lunch where the protein is present.

Event A = Has tumours

Event B = lunch with protein

P(B) = $fracn100$.
P(B') = $frac100-n100$.
P(A|B) = $frac12$. P(A|B') = $frac110$
P(B|A) = $frac0.5n/100((0.5n/100) + (100-n)/(1000)$ = $frac5n4n+100$

Question: Find the largest number of pigs that ought to be given the lunch with the protein present if the second researcher wants the above probability to be smaller than 0.5.

$frac5n4n+100$ < 0.5

n < $frac503$ = 16.7

And so n=16.

The difficulty that I have is to evaluate and explain the probability of the first scenario using the largest number of pigs found in the second scenario.

My comment on this is that:

For an upper bound of 16, the lower the the number of pigs given the lunch the smaller the probability of the pig being given the lunch.

I have provided my working out for the various stages of this question. However I am having real difficulty with the explanation part of the question - can anyone provide with their thoughts, am I missing something here?

Thanks in advance.

In the case $n=16$, we could think about the the expected number of pigs with tumours for each of the lunches:

• Of the $16$ pigs that ate the protein lunch, we expect $8$ of them to get tumours (because the probability of a pig in this group getting a tumour is $0.5$).


• Of the $84$ pigs that ate the normal lunch, we expect $8.4$ of them to get tumours (because the probability of a pig in this group getting a tumour is $0.1$).

This gives some intuition for why a pig with a tumour is more likely to have had the normal lunch (in the case $n = 16$): We expect there to be $16.4$ pigs with tumours, and we expect $8$ of them to have eaten the protein lunch.

If $n < 16$ then

• Of the $n$ pigs that ate the protein lunch, we expect $n/2$ (less than $8$) of them to get tumours.


• Of the $100 - n$ pigs that ate the normal lunch, we expect $(100 - n)/10$ (greater than $8.4$) to get tumours.

And this gives us an intuition for why, when $n < 16$, the probability that a pig with tumours ate the protein lunch is still less than $0.5$.

And in the case $n = 17$ we expect to see $8.5$ pigs who both ate the protein lunch and got tumours. We expect to see $8.3$ pigs who both ate the normal lunch and got tumours.