Answer:
Given the following events and its elements when two 6-sided dice are tossed:
A: the sum of the dice is even
B: at least one die shows a 3
C: the sum of the dice is 7
The elements of the intersections are:
a) A∩B={(1, 3),(3, 1),(3, 3),(3, 5),(5, 3)}
b) B^c∩C={(1, 6),(2, 5),(5, 2),(6, 1)}
c) A∩C={∅}
d) A^c∩B^c∩C^c={(1, 2),(1, 4),(2, 1),(4, 1),(4, 5),(5, 4),(5, 6),(6, 5)}
Step-by-step explanation:
The total number of elements of the universal set (U) for this problem is 36 elements because the number of possible combinations is 6*6.
For the event A, half of the elements satisfy the condition of the sum being an even number.
A={(1, 1),(1, 3),(1, 5),...,(6, 2),(6, 4),(6, 6)}=18 elements
For event B, the elements that contain a 3 are:
B={(1, 3),(2, 3),(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 3),(5, 3),(6, 3)}= 11 elements
For event C, the sum of the elements is 7:
C={(1, 6),(2, 5),(3, 4),(4, 3),(5, 2),(6, 1)}=6 elements
Now let's find the intersections:
a) A∩B are the elements of A that have a 3.
A∩B={(1, 3),(3, 1),(3, 3),(3, 5),(5, 3)}
b) B^c∩C are the elements of the universal set (U) that do not have a 3 and that the sum of the dice is 7
B^c∩C={(1, 6),(2, 5),(5, 2),(6, 1)}
c) A∩C are the elements of that sum 7, but this is not possible given that all the elements of A sum an even number and 7 is not an even number.
A∩C={∅}
d) A^c∩B^c∩C^c are the elements that don't sum an even number, don't have a 3 and the sum is not 7.
A^c∩B^c∩C^c={(1, 2),(1, 4),(2, 1),(4, 1),(4, 5),(5, 4),(5, 6),(6, 5)}
