1. Use the axioms of probability to show the following:
(a) For any event E, P{Ec} = 1 − P{E}.
(b) P{∅} = 0.
(c) For any events E and F, if E ⊂ F, then P{E} ≤ P{F}.
(d) If the events E1, E2, and E3 are mutually exclusive, then P{E1∪E2∪E3} = P{E1}+P{E2}+P{E3}.
can someone get me started on these?
Certainly!
To prove these statements using the axioms of probability, we will mainly rely on the three axioms:
1. Nonnegativity: For any event E, P{E} ≥ 0.
2. Additivity: For any two mutually exclusive events E and F, P{E ∪ F} = P{E} + P{F}.
3. Normalization: The probability of the entire sample space Ω is equal to 1, denoted as P{Ω} = 1.
(a) To prove that P{Ec} = 1 − P{E}, we can use the additivity axiom. Since E and Ec are complementary events (meaning if one occurs, the other does not), we have E ∪ Ec = Ω (the entire sample space). Therefore, applying the additivity axiom, we get:
P{E ∪ Ec} = P{Ω} = 1.
Since E and Ec are mutually exclusive, we can also use the additivity axiom to write:
P{E ∪ Ec} = P{E} + P{Ec}.
As a result, we obtain the equation:
P{E} + P{Ec} = 1.
By isolating P{Ec}, we find:
P{Ec} = 1 − P{E}.
(b) To show that P{∅} = 0, we consider the empty set ∅. Since it does not contain any outcomes, we know that P{∅} = 0. This follows directly from the nonnegativity axiom, as the probability of an empty set should be zero.
(c) For any events E and F, if E ⊂ F (E is a subset of F), we need to prove that P{E} ≤ P{F}.
Consider the event F. Since E ⊂ F, we can write F as a disjoint union of two events: F = E ∪ (F \ E).
Using the additivity axiom, we get:
P{F} = P{E ∪ (F \ E)} = P{E} + P{F \ E}.
Since probabilities cannot be negative, we have P{F \ E} ≥ 0. By rearranging the equation, we find:
P{E} = P{F} - P{F \ E}.
Since P{E} is the sum of two non-negative terms, it follows that P{E} ≤ P{F}.
(d) If the events E1, E2, and E3 are mutually exclusive, we need to show that P{E1 ∪ E2 ∪ E3} = P{E1} + P{E2} + P{E3}.
Using the additivity axiom, we can write:
P{E1 ∪ E2 ∪ E3} = P{(E1 ∪ E2) ∪ E3}.
Since E1 and E2 are mutually exclusive, E1 ∪ E2 is the disjoint union of the two events. Thus, we have:
P{(E1 ∪ E2) ∪ E3} = P{E1 ∪ E2} + P{E3}.
Applying the additivity axiom again, we obtain:
P{E1 ∪ E2} + P{E3} = P{E1} + P{E2} + P{E3}.
Therefore, P{E1 ∪ E2 ∪ E3} = P{E1} + P{E2} + P{E3} for the mutually exclusive events E1, E2, and E3.