SOLUTION: GEmployees have a choice of going out to lunch or eating in the cafeteria. There is a 40% chance that an employee eats lunch in the cafeteria. The probability that a worker arrives

Question 1083271: GEmployees have a choice of going out to lunch or eating in the cafeteria. There is a 40% chance that an employee eats lunch in the cafeteria. The probability that a worker arrives back from lunch on time is .44. The probability that an employee arrives back from lunch on time given that he/she ate in the cafeteria is .6.
What is the probability an employee eats in the cafeteria and arrives late from lunch?
What is the probability an employee does not eat in the cafeteria and does not arrive late from lunch?
What is the probability an employee does not eat in the cafeteria given that he/she arrives on time?
What is the probability an employee eats in the cafeteria or arrives late from lunch?
What is the probability an employee does not eat in the cafeteria given that he/she does not arrive on time?
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
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.
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