Download Probability Theory: Lecture Notes for Math331, Fall 2008 - Foundations by David Anderson and more Study notes Mathematics in PDF only on Docsity!
Section 1.3 lecture notes Math331, Fall 2008 Instructor: David Anderson
Section 1.3: foundations
Definition 1. Let S be a sample space. Suppose that for each A ⊂ S there is a number P (A) that satisfies
- P (A) ≥ 0 for all A.
- P (S) = 1.
- ∗ If {Ai} are mutually exclusive (Ai ∩ Aj = ∅ if i 6 = j), then
P
i=
Ai
∑^ ∞
i=
P (Ai).
Then P is a probability for S and P (A) is the probability of A.
Example 1. Consider flipping a coin and recording the outcome. Then S = {H, T }. Need a probability. Assume fair coin. Then
P {H} = 1/ 2 P {T } = 1/ 2 P {H, T } = 1.
Note that P ({H} ∪ {T }) = 1 = 1/2 + 1/2 = P {H} + P {T }.
This is not the only probability possible for this experiment/sample space. Could have an unfair coin.
P {H} = p P {T } = 1 − p P {H, T } = 1.
Simple Theorems
Theorem 1. P (∅) = 0.
Proof.
P (S) = P (S ∪
⋃^ ∞
i=
∅) = P (S) +
∑^ ∞
i=
P (∅).
Thus, P (∅) = 0.
Theorem 2. Let A 1 ,... , An be mutually exclusive (but finite). Then,
P (
⋃^ n
i=
Ai) =
∑^ n
i=
P (Ai).
Proof. Use previous result and take sum to infinity with the rest being the empty set.
Example: Rolling a dice. P ({ 1 , 2 } ∪ { 3 }) = P ({ 1 , 2 }) + P ({ 3 }) = P ({ 1 }) + P ({ 2 }) + P ({ 3 }) = 1/6 + 1/6 + 1/6 = 1/2.
Corollary 1. For any set A, P (A) ≤ 1.
Proof. P (A ∪ Ac) = P (S) = 1. But, A and Ac^ are mutually exclusive. So, 1 = P (A ∪ Ac) = P (A) + P (Ac). Non-negativity now shows P (A) ≤ 1.
In book with examples. Will be used throughout course:
Theorem 3. Let S be a sample space with N elements that all have same probability of occurring. Then for any A ⊂ S,
P (A) =
{# elements of A} N
Proof. Let A consist of j elements, {w 1 , w 2 ,... , wj } ⊂ S. Then by the sets {wi} being mutually exclusive,
P (A) = P {w 1 , w 2 ,... , wj } = P {w 1 } + P {w 2 } + · · · + P {wj } = j/N.
Example 2. A number is selected at random from the numbers { 1 ,... , 52 }. What is the probability that the number is divisible by 5?
Solution: S = { 1 ,... , 52 } and N = 52. 52/5 = 10 + 2/5 so there are 10 numbers in S that are divisible by 5 (why?). Thus the probability is 10/52.
Caution: You do not always have sample points with equal probability! This could depend upon your choice of sample space. Example: Consider all families with two children. You want to describe the gender of the children. One option which accounts for age is
S = {bb, bg, gb, gg}.
Another, that doesn’t account for gender, would be
W = {bb, gb, gg}.
W does not have equiprobable events. P {gb} = .5.
