SOLUTION: Let A is event a child is infected with ascariasis, and G be event a child infected with giardiasis. Suppose P(A)= .30, P(G)=.25 and P(AnG)=.13. Then what is the probability that

Algebra ->  Probability-and-statistics -> SOLUTION: Let A is event a child is infected with ascariasis, and G be event a child infected with giardiasis. Suppose P(A)= .30, P(G)=.25 and P(AnG)=.13. Then what is the probability that       Log On


   

Question 1175344: Let A is event a child is infected with ascariasis, and G be event a child infected with giardiasis. Suppose P(A)= .30, P(G)=.25 and P(AnG)=.13.
Then what is the probability that s child randomly selected will test negative for these intestinal parasite?
Answer by ikleyn(52780) About Me  (Show Source):
You can put this solution on YOUR website!
.

It seems to me that to be infected is different event as to be tested positively;
as well as to be not infected is different event as to be tested negatively.


So, if in your post/problem you try to use professional terminology, then do it consistently. 


Now, if to forget about these details, the probability to be infected at least one one of "these intestinal parasites" is


    P(A or B) = 0.3 + 0.25 - 0.13 = 0.42.


The probability do not be infected is the COMPLEMENT to it, i.e.


    1 - 0.42 = 0.58.      ANSWER

Solved.