I'll give this a shot. I'm interpreting "predictive value" to mean the estimated probability that the disease is present, given that the symptom is (not) reported.
Let S be be event that the patient reports the symptom, and let D be the event that the patient has the disease.
Based on the data, we can estimate the probability of a patient reporting the symptom given that they have the disease as:
P(S | D) = 744/755
And we can estimate the probability of a patient reporting the symptom given that the patient does not have the disease as:
P(S | ¬D) = 21/1380
We know logically that P(S) = P(S ∧ D) + P(S ∧ ¬D), because these events are mutually exclusive. By Bayes' theorem we have:
P(S ∧ D) = P(D) * P(S | D)
P(S ∧ ¬D) = P(¬D) * P(S | ¬D) = (1 - P(D)) * P(S | ¬D)
And therefore:
P(S) = P(D) * 744/755 + (1 - P(D)) * 21/1380
P(S) = P(D) * 67391/69460 + 21/1380
Again by Bayes' theorem we have for the positive case:
P(D | S) = P(S | D) * P(D) / P(S)
P(D | S) = 68448 / (67391 + (1057 / P(D)))
And the negative case:
P(D | ¬S) = P(¬S | D) * P(D) / P(¬S)
P(D | ¬S) = (1 - P(S | D)) * P(D) / (1 - P(S))
P(D | ¬S) = 1012 / (-67391 + (68403 / P(D)))
Now for the given values of P(D), 0.0001, 0.01, and 0.01, we can compute the respective values of P(D | S): 22816/3656797, 22816/57697, and 22816/25987. Also we can compute the respective values of P(D | ¬S): 92/62178419, 92/615719, and 1012/616639.
My standard method: Imagine a population of 10000 people. Since 10% of the population are sick, there are 1000 sick people, 9000 healthy people. 0.002(900)= 18 of the healthy people test positive and 0.60(1000)= 600 of the sick people test positive, a total of 618 people who test positive.
Of the 618 people who test positive, 18/618= 0.029, 2.9% are actually healthy.
Best Answer
Hint:
Suppose you had $1000$ people.