A test for syphilis gives a positive result with probability 0.98 when tested on a person who has the disease. The test gives a negative result with probability 0.9 when tested on a person who does not have the disease. Assume that 3% of a society has the disease, and compute the following probabilities in the order they are asked. Note that for each individual there are two disease status(disease, no disease) and two types of test results(positive, negative). (a) (0.5 point) Compute the probability that an individual has the disease and has a positive test result. Explain your computation. (b) (0.5 point) Compute the probability

The probability that an individual has the disease and has a positive test result is:

9.8%.

Step-by-step explanation:

Probability of a true positive result for a test of syphilis = 0.98 or 98%

Probability of a false negative result for a test of syphilis = 0.02 or 2% (100 - 98)

Probability of a true negative result for a test of syphilis = 0.9 or 90%

Probability of a false negative result for a test of syphilis = 0.1 or 10% (100 - 90)

Therefore, the probability that an individual has the disease and has a positive test result = 0.98 * 0.1 = 0.098 = 9.8%

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.