Question 1083271
Tree diagram
.4C=======================.6 NO
.6 on time (0.24)=======
.4 late (0.16)==========
This is how I understand the problem.  Then there is an overall probability that a worker arrives back on time is 0.44, which means a 0.20 probability that the worker who eats out arrives on time.  That means that they have a 1/3 probability of returning on time and a 2/3 probability of being late.

With these assumptions.
a. Eats in cafeteria and arrives late is 0.16, 0.4*0.4
b. Does not eat there and is not late 0.20
c.  Given that they arrive on time (44%) probability they didn't eat in cafeteria is 20/44=0.45
d. Eats in cafeteria=0.4+arrives late 0.56- both -0.16=0.80
e. Given that did not arrive on time (0.56) probability they did not eat in the cafeteria (.40) so that is 0.71.