Question 23891
-His first car started 20 times
P(1st car starts)=20/30=2/3; P(it doesn't start)=1/3
-----------
-His second car started 18 times
P(2nd car starts)=18/30=3/5; P(it doesn't start)=2/5
-----------
-Both cars started 40% of the time
P(both start)=2/5
-------------------- 
What is the probability that on any particular morning during the month: 
a) at least one of the cars start?
P(at least one starts)=1-P(none starts)=1-(1/3)(2/5)=1-2/15=13/15
------------------- 
b) he cannot start either of his two cars?
P(neither starts)= 1/3*2/5 = 2/15
Cheers,
Stan H.