A laboratory blood test is 99 % effective in | vedclass

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Question

(A) Let $E_{1}$ be the event that a person has the disease and $E_{2}$ be the event that a person does not have the disease.Given that $P(E_{1}) = 0.1

Vedclass · Accepted Answer

$\frac{22}{133}$

Vedclass · Answer

(A) Let $E_{1}$ be the event that a person has the disease and $E_{2}$ be the event that a person does not have the disease.Given that $P(E_{1}) = 0.1 \% = 0.001$.Since $E_{1}$ and $E_{2}$ are complementary events,$P(E_{2}) = 1 - P(E_{1}) = 1 - 0.001 = 0.999$.Let $A$ be the event that the blood test result is positive.$P(A | E_{1}) = 99 \% = 0.99$ (Probability of positive result given the person has the disease).$P(A | E_{2}) = 0.5 \% = 0.005$ (Probability of positive result given the person is healthy).We need to find $P(E_{1} | A)$. By Bayes' theorem:$P(E_{1} | A) = \frac{P(E_{1}) \cdot P(A | E_{1})}{P(E_{1}) \cdot P(A | E_{1}) + P(E_{2}) \cdot P(A | E_{2})}$Substituting the values:$P(E_{1} | A) = \frac{0.001 	imes 0.99}{(0.001 	imes 0.99) + (0.999 	imes 0.005)}$$= \frac{0.00099}{0.00099 + 0.004995}$$= \frac{0.00099}{0.005985}$$= \frac{990}{5985}$$= \frac{22}{133}$

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