Drinking water may be contaminated by two pollutants. In a given community, the

Question

Drinking water may be contaminated by two pollutants. In a given community, the probability of its drinking water containing an excessive amount of pollutant A is 0. whereas that of pollutant B is 0.2. When pollutant A is excessive, it will definitely cause health problems; however, when pollutant B is excessive, it will cause health problems in only 20% of the population who have a low natural resistance to that pollutant. Also, data from many similar communities reveal that the presence of these two pollutants is not independent; half of these communities whose drinking water contains excessive amounts of pollutant A will also contain excessive amounts of pollutant B Suppose a resident is randomly selected from this community, what is the probability that he or she will suffer health problems from the drinking water?

Explanation / Answer

Let in a given community, drinking water contaminated by two pollutants. probability that drinking water contaminated by pollutant A is 0.1 and pollutant B is 0.2 respectively.

P(A) = 0.1 = 10%

P(B) = 0.2 = 20%

given that A and B are not independent also half of these communities whose drinking water contains excessive amounts of pollutant A will also contain excessive amount of pollutant B this means probabability of resident experience these pollutant is 1/2 therefore,

probability of randomly selected resident suffer health problem = (1/2) *P(A)* P(B)

= 0.5 *0.1*0.2

= 0.01

probability of randomly selected resident suffer health problem = 0.01

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