Question

An experiment consists of tossing a single die and observing the # of dots that show on the upper face
A:observe a # less than 4
B: Observe a # less than or equal to 2
C: observe a # greater than 3
Find the probabilities for:
1. A(intersection)B
2. A(intersection)B(intersection)C
3 AUC

Answer 1

1. Because B is a subset of A, A intersect B is just B. So P(A intersect B) = P(B) = 1/3
2. Again, A and B intersected is B, so this is the same as B intersect C. B and C are mutually exclusive events (a number cannot be both less than or equal to 2 AND greater than 3), so the event is impossible. Thus its probability is 0.
3. A union C means either or. All numbers a less than 4 or greater than 3, so the probability is 1.

Answer 2

@alexray19 is correct

Answer 3

i dont understand 3 though ):

Answer 4

Event A is the number is less than 4, and event C is a number is greater than 3. If you think about it, all numbers will fall into at least one of these two events. That is, every number is either less than 4 or greater than 3. When you union two events, you're saying you'll accept anything that falls into either of the events - it's an OR operation. In this case, all numbers do, so the probability is 1.

Answer 5

thanks so much :)