Question

Find the probability that a person taking vaccine $A_2$ will get infected.

Answer 1

Concept:
This problem is based on Conditional Probability and the Law of Total Probability. Let \[ A_1=\text{Person received Vaccine }A_1,\qquad A_2=\text{Person received Vaccine }A_2,\qquad A_3=\text{Person received Vaccine }A_3 \] Let \[ P=\text{Person is protected from infection} \] and \[ I=\text{Person gets infected}. \] Since a person is either protected or infected, \[ P(I)=1-P(P). \] Given, \[ P(A_1)=0.25,\qquad P(A_2)=0.35,\qquad P(A_3)=0.40 \] Also, \[ P(P|A_1)=0.60,\qquad P(P|A_2)=0.55,\qquad P(P|A_3)=0.50. \] Therefore, \[ P(I|A_1)=1-0.60=0.40, \] \[ P(I|A_2)=1-0.55=0.45, \] \[ P(I|A_3)=1-0.50=0.50. \] 

Since \[ P(I|A_2)=1-P(P|A_2), \] we have \[ P(I|A_2)=1-0.55=0.45. \] Hence, \[ \boxed{P(I|A_2)=0.45} \] or \[ \boxed{\frac{9}{20}}. \] 

Answer 2

Using the Law of Total Probability, \[ P(P)=P(A_1)P(P|A_1)+P(A_2)P(P|A_2)+P(A_3)P(P|A_3). \] Substitute the given values: \[ =0.25(0.60)+0.35(0.55)+0.40(0.50). \] Calculate each term: \[ =0.15+0.1925+0.20. \] Therefore, \[ P(P)=0.5425. \] Hence, \[ \boxed{P(P)=0.5425} \] or \[ \boxed{\frac{217}{400}}. \]

Answer 3

Using Bayes' Theorem, \[ P(A_1|I)=\frac{P(A_1)\,P(I|A_1)}{P(I)}. \] First find \[ P(I)=1-P(P). \] Since \[ P(P)=0.5425, \] we get \[ P(I)=1-0.5425=0.4575. \] Now, \[ P(A_1|I)=\frac{0.25\times0.40}{0.4575}. \] \[ =\frac{0.10}{0.4575} =\frac{40}{183} \approx0.2186. \] Hence, \[ \boxed{P(A_1|I)=\frac{40}{183}\approx0.219.} \]

Answer 4

A person is not infected means the person is protected. Using Bayes' Theorem, \[ P(A_3|P)=\frac{P(A_3)\,P(P|A_3)}{P(P)}. \] Substitute the values: \[ =\frac{0.40\times0.50}{0.5425}. \] \[ =\frac{0.20}{0.5425} =\frac{80}{217} \approx0.369. \] Hence, \[ \boxed{P(A_3|P)=\frac{80}{217}\approx0.369.} \]