Question

In: Statistics and Probability

In a certain urban area the probability that a middle-age person has hypertension is 0.2.

In a certain urban area the probability that a middle-age person has hypertension is 0.2. The probability that a middle-age person has high cholestrol is 0.8. Assume that these two conditions are mutually independent. The probability that a randomly chosen middle-age person suffers from at least one of these two conditions is closest to:

Solutions

Expert Solution

it is given that the two event are mutually independent .

Let P(A) be the probability that person has hypertension and P(A') be the probability that person has not hypertension, so P(A) = 0.2 and P(A') = 1-0.2 = 0.8

Let P(B) be the probability that person has high cholestrol and P(B') be the probability that person has not high cholestrol, so P(B) = 0.8, and P(B') = 1-0.8 = 0.2

now we have to find the probability that a randomly choosen person suffer from atleast one of the two condition

so cases for support of our problem will be

1: P(A and B') that person has hypertension but not high cholestrol

2) P(A' and B) that person has not hypertension but has high cholestrol

3) P(A and B) that person is suffered by both hypertension and high cholestrol

now, P(A and B') = 0.2*0.2 = 0.4

P(A' and B) = 0.8*0.8 = 0.64

P(A and B) = 0.2*0.8 = 0.16

so required probability = 0.4+0.64+0.16 = 0.84


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