4) There are several chemicals which are well known to cause cancer in humans. L

Question

4) There are several chemicals which are well known to cause cancer in humans. Let us consider three such chemicals which have recently been shown (see http://news.bbe.co.uk/I/hi/sci/tech/1098564.stm, for example) to have some chance of occurring in salmon, particularly in farmed salmon: PCB's, chlorine pesticides (CP's), and polybrominated diphenyl ethers (PDE's). A vastly simplified and hypothetical cancer risk assessment due to eating a single salmon might proceed as follows: Assume that a salmon has probabilities 30%, 50%, and 20% of containing PCB's, CP's, and PDE's, respectively. Furthermore, assume that there is a 5% chance of containing both PCB's and CP's, a 5% chance of containing both PCB's and PDE's, a 2% chance of containing both CP's and PDE's and a 1% chance of containing all three chemicals. Suppose that there is a 001% chance of developing cancer due to ingesting a salmon which contains PCB's alone, a 0.005% chance due to ingesting a salmon which contains CP's alone, and a 0.002% chance due to ingesting a salmon containing PDE's alone. If a salmon containing two of the chemicals is ingested, the cancer risk nses to 005%, while if all three chemicals are present in the ingested salmon, the risk becomes 0.1%. (NOTE: a real risk assessment would be considerably more complicated than this.) a) If a person eats a salmon, what is his or her risk of developing cancer? b) If a person does develop cancer from ingesting a particular salmon, what is the probability that the salmon only contained CP's? c) If a person does develop cancer from ingesting a particular salmon, what is the probability that the salmon contained two or more of the chemicals?

Explanation / Answer

P(PCB) = 0.3, P(CP) = 0.5 and P(PDE) = 0.2
P(PCB and CP) = 0.05
P(PCB and PDE) = 0.05
P(CP and PDE) = 0.02
P(PCB and CP and PDE) = 0.01

P(C|PCB) = 0.0001
P(C|CP) = 0.00005
P(C|PDE) = 0.00002

P(C|(PCB and CP)) = P(C|(PCB and PDE)) = P(C|(PDE and CP))= 0.0005
P(C|(PCB and CP and PDE)) = 0.001

(A)
P(C) = P(C|PCB)*P(PCB) + P(C|CP)*P(CP) + P(C|PDE)*P(PDE) + P(C|(PCB and CP))*P(PCB and CP) + P(C|(PCB and PDE))*P(PCB and PDE) + P(C|(PDE and CP))*P(CP and PDE) + P(C|(PCB and CP and PDE))*P(PCB and CP and PDE)

P(C) = 0.3*0.0001 + 0.5*0.00005 + 0.2*0.00002 + 0.0005*0.05 + 0.0005*0.05 + 0.0005*0.02 + 0.01*0.001 = 0.000129

(B)
P(CP|C) = P(C|CP)*P(CP) / P(C) = 0.00005*0.5 / 0.000129
= 0.1938

(C)
Required probability = (P(C|(PCB and CP))*P(PCB and CP) + P(C|(PCB and PDE))*P(PCB and PDE) + P(C|(PDE and CP))*P(CP and PDE) + P(C|(PCB and CP and PDE))*P(PCB and CP and PDE)) / P(C)

= (0.0005*0.05 + 0.0005*0.05 + 0.0005*0.02 + 0.01*0.001)/0.000129 = 0.5426

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