Since we tossed 2 dice, the sample space : 6 x 6 = 36 outcomes: {(1,1), (1,2),(1,3),(1,4),(1,5),(1,6) {(1,2),(2,2),(3,2),(4,2),(5,2)),(6,2) {------------------------------------------ ----------------------------------------- {(1,6),(2,6),(3,6),(4,6),(5,6),(6,6)}
What are the number that their sum is equal to 5: 1+4 =5 4+1 =5 2+3 =5 3+2 =5
Total favorable outcomes = 4:
So the probability of getting a sum of 5 is P(sum=5) = 4/36 = 1/9 = 0.111
That is, the union has as many members as the sum of the number of members of the individual sets, minus the number of elements contained in both sets (to avoid double-counting).
Therefore, will have the most elements when the sets and are disjoint, i.e. , which would mean the most we can can in this case would be