Question

A particular genetic trait occurs in 2% of the population. The reliability of a test to discover the trait is: if the person has the trait then the test is positive 98% of the time, but the test is also positive 3% of the time for those that do not have the
trait.
a)Construct a tree diagram to represent the population and the testing reliability for the genetic trait
b)if a person is selected at random, what is the probability that the person has the trait, given that he tested positive?

Please verify if my tree diagram is correct:
/positive- have trait
-do not have trait
a)population- trait
-no trait
\negative
/ \
have trait do not have trait
b)0.2x0.98=0.196
Thanks in advance

Answer 1

A tree diagram includes the related probabilities, which is the real purpose of the tree diagram, because it then enables us to get the probability of each situation.

            /positive(0.98*0.02=0.0196)
    trait(0.02)-- negative(0.02*0.02=0.0004)
/
\
    no trait(0.98) -- positive (0.03*0.98=0.0294)
            \ negative(0.97*0.98=0.9506)
Sum of all probabilities=1.0 √

For part (b), it is a conditional probability, so the sample space is among those that test positive, namely, 0.0196+0.0294=0.0490.
Out of these, 0.0196 has the trait, so P(T|+)=0.0196/0.0490

Answer 2

Thank you very much for you help and explanations.

Answer 3

You're welcome!