A certain virus infects one in every 300 people. A test used to detect the virus in a person is positive 80% of the time if the person has the virus and 10% of the time if the person does not have the virus. (This 10% result is called a false positive.) Let A be the event "the person is infected" and B be the event "the person tests positive". DRAW A TREE DIAGRAM IN YOUR NOTES AND USE IT TO HELP YOU SOLVE THIS PROBLEM. Find the probability that a person has the virus given that they have tested positive; i.e. find P(AIB). Give your answer as a decimal number and include at least 3 or more non-zero digits. P(AIB)=

In the tree, the first branch will be person has virus or person doesn't have the virus.P(virus) = 1/300P(not virus) = 299/300Now,Then we branch out from each option. These branches would be positive or negative.If they have virus:P(positive) = 0.8P(negative) = 0.2If don't have virus:P(positive) = 0.1P(negative) = 0.9Now, solving the question of probability that a person has the virus given that they have tested positive:We find:P(A|B)P(has virus | positive test) = P(positive and has virus) / P(positive test)P(positive and has virus) = 4/5 * 1/300 = 4/1500P(positive test) = 1/300 * 4/5 + 1/10 * 299/300= (4/1500)+(299/3000)=(8/3000) + (299/3000) = 307/3000= 0.10233So,P(positive and has virus) / P(positive test) = 4/1500 divided by 299/3000 = 0.02675