Two 6-sided dice are tossed. one die is red and the other is white, so that they are distinguishable. (that is, we consider the pair of numbers (number showing on red die, number showing on white die), for example, one outcome is (3, 1), which is different from (1, let a be the event that the sum of the dice is even, b be the event that at least one die shows a 3 and c be the event that the sum of the dice is 7. identify the elements of: (a) a intersection b (b) b^c intersection c (c) a intersection c (d) a^c intersection b^c intersection c^c

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)}

hello from mrbilldoesmath!

answer:

a(n) - a(n-1) -3

discussion:

this ones a bit interesting. given a(n) = 5 - 3n consider this:

a(n-1) = 5 - 3(n-1)

          = (5 - 3n) + 3             => but a(n) = 5 - 3n, the first term

          = a(n)   + 3                 => subtract 3 from both sides

a(n-1) -3 = a(n)  

or

a(n) - a(n-1) -3

you,

mrb

6 meters

step-by-step explanation: