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The problem is: 95 people participated in an experiment. $N$ people were given food with an enzyme in it to help people lose weight. However, some people have gained an illness from eating the food with enzymes in it. It is stated that the probability of a person developing the illness can increase from $0.1$ with normal food to $0.5$ with the enzyme.

i) what is the probability, if the person has an illness, to have eaten the food with the enzyme?

ii) What is the maximum N so that the probability in i) is less than 0.65

I assume that the information given is that total is 95 $$ P(\text{illness} \mid \text{normal}) = 0.1 $$ $$ P(\text{illness} \mid \text{enzyme}) = 0.5 $$

I believe in part i) I need to use Bayes' Theorem so I do this:

$$ P(\text{enzyme} \mid \text{illness}) = \frac{P(\text{illness} \mid \text{enzyme}) \times P(\text{enzyme})}{P(\text{illness})}, $$

but I am not sure how to get the other probabilities to work that out. I could really use the help

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    $\begingroup$ I would recommend drawing a table. That will make things very clear. To continue, you need P(E) as a function of N, then P(I) will law of total probability. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 12 at 22:20
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$P(\text{enzyme})$ is N/95 and N is something to be adjusted.

For $P(\text{illness})$ you use the very useful breakdown for the denominator in Bayes' theorem $P(B)=P(B\mid A)~P(A) + P(B\mid \overline A)~(1-P(A))$, here $$P(\text{illness}\mid \text{enzyme})~P(\text{enzyme})+P(\text{illness}\mid\text{normal})~(1-P(\text{enzyme}))$$

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  • $\begingroup$ PS: This is known as the Law of Total Probability. $\endgroup$ – Graham Kemp Feb 12 at 23:38

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