Question 842382
P(1) = P(2) = ... = P(6) = 0. For each problem, find the number of admissible cases and divide by 6.


For example, #3:
P(odd number or a number greater than 4):
Admissible numbers: 1, 3, 5, 6, each with 1/6 probability
P(odd number or a number greater than 4) = 4/6 = 2/3