Answer:
P(A) = 1/6
P(B) = 1/4
P(A∩B) = 1/6
Explanation:
First let's compute the sample space for all possible sums
For each number that rolls on the six sided die, there are four possible numbers that can roll on the four-sided die
For example if the six-sided die comes up with 1 then the possibilities are i
(1,1), (1, 2), (1, 3) , 1,4)
That is a total of 4 possible outcomes for one number on the six-sided die
So total number of outcomes possible for all 6 numbers rolled on die one and all 4 numbers rolled on die 2 = 6 x 4 = 24 possible outcomes each of them resulting in a sum of the numbers rolled
Let's figure out the number of ways you can get a sum of 5
Six-sided Four-sided
1 4
2 3
3 2
4 1
We can define the set A as { (1,4), (2, 3), (3, 2), (4, 1)} which has 4 possibilities
So if A is the outcome that the sum is 5
P(5) = P(A) = 4/24 = 1/6
Let's figure the number of ways you can get 9 as a sum
Six-sided Four-sided
5 4
6 3
Only two possibilities because the second die can roll only 1, 2, 3 4
Total 2 possibilities: {(4, 5), (6, 3)}
P(9) = 2/24 = 1/12
If B is the event that either a 5 or a 9 is rolled event that would be:
P(5 U 9} = P{A U 9} = P(A) + P(B) = 1/6 + 1/12 = 2/12 = 1/4
So P(B) = 1/4
The set B has 4 + 2 = 6 possible outcomes(either sum = 5 or sum = 9)
{ (1,4), (2, 3), (3, 2), (4, 1), (5, 4), (6, 3)}
For P(A ∩ B) . Let's find the set {A ∩ B}
We have
A = {(1,4),(2,3),(3,2),(4,1)}
B = { (1,4) , (2, 3), (3, 2), (4, 1), (5, 4), (6, 3)}
A ∩ B is the set of elements common to both sets.
This would be just A = {(1,4),(2,3),(3,2),(4,1)}
P(A∩B) = P(A) = 1/6
This is because B is a superset of A and contains all the elements of A