Probability Question: 2. The probability that a test detects the disease if a person has the disease is 0.95..?

Cassidy Pangrazio: one way: suppose the population is 10000.5% = .005(10000) = 50 people will have the disease and 9950 will notP(false positive) = 1 - .90 = .10 = P(detecting disease in person who does not have it)since 9950 people do not have the disease, .1(9950) = 995 people who do not have the disease will test positive50 people do have the disease; thus, 95(50) = .475 people will test positive who do have the diseasethus, 995.475 out of 10,000 will test positiveof these, only .475 actually have the diseaseP(disease | positive test) = .475 / 995.475 = .00048this test isn't much good as a predictor, since it has a .10 (10%) rate of false positives!...Show more

Jarrod Darnall: X: take fee a million if that individual have illness and 0 otherwise. Y: take fee a million if device tell "sure, u are a zombie, shelter!!!!" and 0 otherwise. we've: *P( Y=a million | X=a million) = .ninety 9 *P( Y=0 | X=0) = .ninety 8 *P( X=a million) = .05 P( X=a million | Y=a million)! = P( X=a million & Y=a million) / P( Y=a million) (a million) P( X=a million & Y=a million) = P( X=a million)* P( Y=a million | X=a million) = .05 * .ninety 9 = .0495 (2) P( Y=a million) = P(X=0) * P(Y=a million | X=0) + P(X=a million) * P( Y=a million | X=a million)= (a million - .05) * (a million- .ninety 8) + .05 * .ninety 9 = .0685 (3) (a million,2,3)=> P(X=a million | Y=a million)= .7226 I dont understand if i pass over something or no longer however the hint look no longer needed....Show more