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When the health department tested private wells in a county for two impurities commonly found in drinking water, it found that 10% of the wells had neither impurity, 90% had impurity A, and 20% had impurity B. (Obviously, some had both impurities.) If a well is randomly chosen from those in the county, find the probability distribution for Y, the number of impurities found in the well.

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Answer:

P(Y= 0) = 0.1

P(Y= 0) = 0.7

P(Y= 0) = 0.2

Step-by-step explanation:

Let Y be number of impurities that can be found in the well,

Let A denote the event that impurity A is randomly found in the well

Here Y can have three values i.e 0 , 1 and 2

✓It will take take the value of 0 when there is no impurity found in the well

✓It will take the value of 1 when when exactly one impurity vis found in the well

✓It will take the value of 2 when when both impurities vis found in the well

CHECK THE ATTACHMENT FOR DETAILED EXPLATION

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