Question

1. The probability that a person selected at random from a population will exhibit the classic symptom of a certain disease is 0.2, and the probability that a person selected at random has the disease is 0.23. The probability that a person who has the symptom also has the disease is .18. A person selected at random from the population does not have the symptom. What is the probability that the person has the disease? 2. In a certain population of women 4 percent have had breast cancer, 20 percent are smokers, and 3 percent are smokers and have had breast cancer. A woman is selected at random from the population. What is the probability that she has had breast cancer or smokes or both?

Question 5. (20 pts.) 1. The probability that a person selected at random from a population will exhibit the classic symptom of a certain disease is 0.2, and the probability that a person selected at random has the disease is 0.23. The probability that a person who has the symptom also has the disease is .18. A person selected at random from the population does not have the symptom. What is the probability that the person has the disease? 2. In a certain population of women 4 percent have had breast cancer, 20 percent are smokers, and 3 percent are smokers and have had breast cancer. A woman is selected at random from the population. What is the probability that she has had breast cancer or smokes or both?

Answer 1

Sol Anso Given data 20 0.23 of Disease = Probability of Symptoms CP(s) Probability that a person Selected at random has the disease Probability that a person who has the symptoms also has the disfase= ㅔ 0.18 So First we arrange the joint probability table from given in formation Symptoms 0:18 (given) No Symptoms Total 0.23 Disease 023-0:18 0.05 No Disease 0.2-0.18=0.02 0.8-0.05 -0.075 1-0.23 - 077 Total 0.2 1-0.2= 0.8 1 So According to our table Probability of [ Disease / No Symptoms) - 0.05 Probability of [ No Disease / No symptoms)] = 0.675 Probability of [No Symptoms] = 0.8

So probability that person has the disease given that ( does not have any symptoms) P= Probability of[disease 8 No symptoms of [ No symptoms] Probability = 0.05 0.0625 0.80 P = 0.0695 Given that women ham had breast cancer = 4% = 0.04 Women Care Smokers 209, = 0.2 women [are smokers & have had Breast Can CE2] = 34. 3y = 0:03 So Joint probability table Breast Cancer Nost Breast Cancer Smokers total 0.03(given) 0.20 0.2 - 0:03 0:17 Total 0.01 Non-Smokers 0.04 -0.03 -0.01 0,20 = 0.80 -0.01 -0.799 =0.80 Total 004 ke 0:17 +0.79 -0.96 1

So According to our tablet Method - I Probabilsity of [smokers + Breast Cancer J= 0.03 Probability of [ Non-smokers + Breast cancer] = 0.01 Probability of [ Smokers + Not Breast Cancer - 0:17 a women So probability that has had [ Breast Cancer OR Smokers or both] 0.03 +0.01 +0.17 021 Method -I If we find the probab bility of [Non-Smokers or Not Breast Cuncer ] It is find in table by us P(Non-smokers 0.79 of Not Breast Cancer) The → P[ Smokers + Breast Cancer +(Both)]: 1-PC Non Smokers / Not Borost Cancer) p=1-0.79 = 0.21