6. Determine all joint probabilities listed below from the following information

Question

6. Determine all joint probabilities listed below from the following information: P(A) 0.7, P(AC) 0.3, P(BIA) 0.4, P(BIAC) P(A and B) = P(A and not B)- P(not A and not B) 0.8 P(A' and B) 7. Bad gums may mean a bad heart. Researchers discovered that 81% of people who have suffered a heart attack had pe- riodontal disease, an inflammation of the gums. Only 30% of healthy people have this disease. Suppose that in a certain com- munity heart attacks are quite rare, occurring with only 15% probability. A. If someone has periodontal disease, what is the probabil- ity that he or she will have a heart attack? Probability = B. If 40% of the people in a community will have a heart attack, what is the probability that a person with periodontal disease will have a heart attack? Probability

Explanation / Answer

Ans:

6)

i)P(A and B)=P(B/A)*P(A)=0.4*0.7=0.28

ii)P(A and not B)=P(A)-P(A and B)=0.7-0.28=0.42

iii)Now,

P(B)=P(B/A)*P(A)+P(B/Ac)*P(Ac)=0.4*0.7+0.8*0.3=0.52

P(A' and B)=P(B)-P(A and B)=0.52-0.28=0.24

iv)As,

P(not A)=P(not A and B)+P(not A and not B)

P(not A and not B)=0.3-0.24=0.06

7)

P(disease/heartattack)=0.81

P(disease/not heartattack)=0.30

P(heartattack)=0.15

a)

P(disease)=0.81*0.15+0.30*(1-0.15)=0.3765

P(heartattack/disease)=P(disease/heartattack)*P(heartattack)/P(disease)

=0.81*0.15/0.3765=0.3227

b)P(heartattack)=0.4

P(disease)=0.81*0.4+0.30*(1-0.4)=0.504

P(heartattack/disease)=P(disease/heartattack)*P(heartattack)/P(disease)

=0.81*0.4/0.504=0.6429

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