Write out all triples that sum to 5:

(1,1,3), (1,3,1), (3,1,1)

(1,2,2), (2,1,2), (2,2,1)

There are no other triples that sum to 5. How many triples are possible? There are six possible rolls for each die. So, there are \(\displaystyle 6^3\) possible triples. Hence, \(\displaystyle P(T) = \dfrac{6}{6^3} = \dfrac{1}{36}\). Then, \(\displaystyle P(J|T)\) is the number of outcomes (among the six triples whose sum is five) that have a one in the third position. I count three. So, \(\displaystyle P(J|T) = \dfrac{3}{6} = \dfrac{1}{2}\).