Tentukan himpunan x-y=2 x+y=1
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Answer:
Step-by-step explanation:
Let's call D the event that a person has the disease, D' the event that a person doesn't have the disease and T the event that the person tests negative for the disease.
So, the probability P(D/T) that a randomly chosen person who tests negative for the disease actually has the disease is calculated as:
P(D/T) = P(D∩T)/P(T)
Where P(T) = P(D∩T) + P(D'∩T)
So, the probability P(D∩T) that a person has the disease and the person tests negative for the disease is equal to:
P(D∩T) = (1/1000)*(0.005) = 0.000005
Because 1/1000 is the probability that the person has the disease and 0.005 is the probability that the person tests negative given that the person has the disease.
At the same way, the probability P(D'∩T) that a person doesn't have the disease and the person tests negative for the disease is equal to:
P(D'∩T) = (999/1000)*(0.99) = 0.98901
Finally, P(T) and P(D/T) are equal to:
P(T) = 0.000005 + 0.98901 = 0.989015