It is known that there only is 1% chance of getting a disease. a test is being devised to detect the disease. the probability that the test comes out positive if the patient has the disease is 0.98 and the probability that the test comes out negative if the patient does not have the disease is 0.95, a. what is the probability that a random test comes out negative? b. what is the probability that a patient really has the disease, if the test came out positive?

Suppose is the event that a given patient has the disease, and is the event of a positive test result.

We're given that





where denotes the complement of an event .

a. We want to find . By the law of total probability, we have



That is, in order for to occur, it must be the case that either also occurs, or does. Then from the definition of conditional probability we expand this as



so we get



b. We want to find . Now, we can use Bayes' rule, but if you're like me and you find the formula a bit harder to remember, we can easily derive it.

By the definition of conditional probability,



We have the probabilities of / occurring given that / occurs, but not vice versa. However, we can expand the probability in the numerator to get a probability in terms of being conditioned on :



Meanwhile, the law of total probability lets us rewrite the denominator as



or in terms of conditional probabilities,



so that



which is exactly what Bayes' rule states. So we get

X^2 - 4x +(4/2^2) + y^2 + 8y + (8/2^2) = -4 + (4/2^2) + (8/2^2)(x + 2)^2 + (y + 4)^2 = -4 + 4 + 16your center is (-2, -4) and the radius is the sqrt of -4 + 4 + 16 which is 4

You would need a laid out graph for this but could you provide a picture of just a graph? i can give stepssolve for y.subtract 4xdivide by -1 (opposites)graph slope intercept formcheck each point to see if it’s on the graph