Question

Two dice are rolled. E is the event that the sum is even, F is the event of rolling at least one six, and G is the event that the sum is eight. List the outcomes for the following events:

a. E ∩ F {(2, 2), (4, 4), (6, 6)} {(6, 2), (6, 4), (6, 6), (2, 6), (4, 6)} {(2, 6), (4, 6), (6, 6)} ∅
b. Ec ∩ G {(6, 2), (6, 4), (6, 6), (2, 6), (4, 6)} {(2, 6), (4, 6), (6, 6)} {(2, 2), (4, 4), (6, 6)} ∅

Answer 1

Answer:

(a)$E \cap F ={(2, 6),(4, 6),(6, 2),(6, 4),(6, 6)}$

(b) $E^c \cap G =\{ \}$

Step-by-step explanation:

The sample space of two dice rolled is given below:

$\{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)\\(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)\\(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)\\(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)\\(5, 1), (5, 2), (5, 3), (5, 4), (5, 5),(5,6)\\(6, 1), (6, 2), (6, 3), (6, 4), (6,5), (6,6)\}$

For Event E (The sum is even), the outcomes are:

$(1, 1), (1, 3), (1, 5),(2, 2), (2, 4), (2, 6)\\(3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6)\\(5, 1),(5, 3), (5, 5), (6, 2), (6, 4), (6,6)$

For Event F (Rolling at least one six), the outcomes are:

$(1, 6), (2, 6), (3, 6), (4, 6),(5,6),(6, 1), (6, 2), (6, 3), (6, 4), (6,5), (6,6)$

For Event G (The sum is eight), the outcomes are:

$(2, 6), (3, 5),(4, 4), (5, 3),(6, 2)$

(a)$E \cap F$

Therefore:

$E \cap F ={(2, 6),(4, 6),(6, 2),(6, 4),(6, 6)}$

(b)$E^c \cap G$

E is the event that the sum is even

Therefore: $E^c is the event that the sum is odd.$

Since G is the event that the sum is eight( which is even), the intersection of the complement of E and G will be empty.

Therefore:

$E^c \cap G =\{ \}$