A laboratory blood test is 99 % effective in | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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}$

Similar Questions

$A$ letter is known to have arrived by post either from $KANPUR$ or from $ANANTPUR$. On the envelope,just two consecutive letters $AN$ are visible. The probability that the letter came from $ANANTPUR$ is:

Box $I$ contains $30$ cards numbered $1$ to $30$ and Box $II$ contains $20$ cards numbered $31$ to $50$. $A$ box is selected at random and a card is drawn from it. The number on the card is found to be a non-prime number. The probability that the card was drawn from Box $I$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module