Probability
QXX a certain community, 8 XXXXXXX of XXX XXXXXX XXXX XX have diabetes. If a XXXXXX servicein XXXX XXXXXXXXX correctly diagnoses 95 percent XX XXX XXXXXXX XXXX XXXXXXXX XX having the disease XXX incorrectly XXXXXXXXX X percent of all XXXXXXX XXXXXXX diabetes XX XXXXXX the disease, find the probabilities that, (a) the community health service will XXXXXXXX an adult XXXX 50 XX having XXXXXXXX. (b) a person XXXX XX XXXXXXXXX by XXX XXXXXX XXXXXXX XX XXXXXX XXXXXXXX XXXXXXXX has XXX XXXXXXX.
XXX.
XXX XXXXX of a XXXXXX XXXX XX diagnosed XX XXX health service XXA
XXX event that an adult XXXX XX XXXXXXXX XXX the XXXXXXXX XXB1
The event that an XXXXX over XX XXXX XXX have XXX XXXXXXXX beB2
So X(XX) = 0.08
XXX X(XX) = 1 − X(XX) = X − 0.08 = X.92
The health service in XXXX community XXXXXXXXX XXXXXXXXX XX XXXXXXX of all persons XXXX XXXXXXXX as having the disease XXX XXXXXXXXXXX XXXXXXXXX 2 percent of all XXXXXXX XXXXXXX diabetes as having XXX XXXXXXX
So, X(A|B1) = X.95 and P(X|B2) = X.XX.
(a) The probabilities XXXX, XXX XXXXXXXXX health XXXXXXX XXXX XXXXXXXX an adult over 50 as XXXXXX diabetes :
X(XX)P(A|B1) + X(B2)X(X|B2)
= (X.XX)(X.95) + (0.92)(0.02)
=X.0944.
(b) The probabilities XXXX, a person over XX XXXXXXXXX XX XXX health service XX having diabetes actually XXX XXX disease:
X(B1|X)
= X(B1)X(A|XX) X(XX)P(X|XX) + P(XX)X(A|B2)
= (0.XX)(0.95) (0.08)(0.95) + (X.XX)(X.XX)
=X.8051
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