The probability that a subject has a certain stomach bacteria is 0.70, the proba
ID: 3201981 • Letter: T
Question
The probability that a subject has a certain stomach bacteria is 0.70, the probability that a subject is showing signs of gastritis is 0.15, and the probability that a subject both has the bacteria and shows signs of gastritis is 0.10.
(a) Find the probability that a subject either shows signs of gastritis, or has the stomach bacteria, or both (the union).
(b) If a subject has the stomach bacteria, what is the probability that they show signs of gastritis?
(c) If a subject does not have the stomach bacteria, what is the probability that they show signs of gastritis?
(d) If a subject shows signs of gastritis, what is the probability they have the stomach bacteria?
(e) Are the events of having the stomach bacteria and showing signs of gastritis independent?
Explanation / Answer
P(B) = 0.7, P(G) = 0.15, P(B and G) = 0.10
(a) P(B or G) = P(B) + P(G) - P(B and G) = 0.70 + 0.15 - 0.10 = 0.75
(b) P(G | B) = P(G and B)/P(B) = 0.10/0.70 = 1/7
(c) P(G | B') = P(G and B')/P(B') =0.05 /0.30 = 1/6
(d) P(B | G) = P(G and B)/P(G) = 0.10/0.15 = 2/3
(e) P(B | G) = 2/3 and P(B) = 0.70 are unequal. So the events are not independent.
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