Answer: 0.31 or 31%
Let A be the event that the disease is present in a particular person
Let B be the event that a person tests positive for the disease
The problem asks to find P(A|B), where
P(A|B) = P(B|A)*P(A) / P(B) = (P(B|A)*P(A)) / (P(B|A)*P(A) + P(B|~A)*P(~A))
In other words, the problem asks for the probability that a positive test result will be a true positive.
P(B|A) = 1-0.02 = 0.98 (person tests positive given that they have the disease)
P(A) = 0.009 (probability the disease is present in any particular person)
P(B|~A) = 0.02 (probability a person tests positive given they do not have the disease)
P(~A) = 1-0.009 = 0.991 (probability a particular person does not have the disease)
P(A|B) = (0.98*0.009) / (0.98*0.009 + 0.02*0.991)
= 0.00882 / 0.02864 = 0.30796
*round however you need to but i am leaving it at 0.31 or 31%*
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Answer:
me too bro me too........
Answer:
A
Step-by-step explanation:
Substitute 6 into place and solving the answer gives you 4 which is indeed bigger than 3. All the other numbers will give you an answer of something lower than 3 which is not what we're looking for in this case.
Answer:
4 : 5
Step-by-step explanation:
8 : 10
4 : 5
hope this was helpful : )
If 4600 corresponds with 100, multiply 4600 by .09 and you get 414.
Also:
46 * 9 = 414
