18.05 Spring 2005 Lecture Notes 18.05 Lecture 1 February 2, 2005
Required Textbook - DeGroot & Schervish, "Probability and Statistics," Third Edition Recommended Introduction to Probability Text - Feller, Vol. 1
1.2-1.4. Probability, Set Operations. What
18.05 Lecture 2 February 4, 2005
1.5 Properties of Probability. 1. P(A) [0, 1] 2. P(S) = 1 3. P(Ai ) = P (Ai ) if disjoint Ai Aj = , i = j The probability of a union of disjoint events is the sum of their probabilities. 4. P(), P(S) = P(S ) = P(S) + P() =
18.05 Lecture 3 February 7, 2005
n! Pn,k = (n-k)! - choose k out of n, order counts, without replacement. nk - choose k out of n, order counts, with replacement. n! Cn,k = k!(n-k)! - choose k out of n, order doesn't count, without replacement.
1.9 Multino
18.05 Lecture 4 February 11, 2005
Union of Events P(A1 . An ) =
i
P(Ai )
i<j
P(Ai Aj ) +
i<j<k
P(Ai Aj Ak ) + .
It is often easier to calculate P(intersections) than P(unions)
Matching Problem: You have n letters and n envelopes, randomly stu the lette
18.05 Lecture 5 February 14, 2005
2.2 Independence of events. P(A|B) = P(AB) ; P(B) Definition - A and B are independent if P(A|B) = P(A) P(A|B) = P(AB) = P(A) P(AB) = P(A)P(B) P(B)
Experiments can be physically independent (roll 1 die, then roll another
18.05 Lecture 6 February 16, 2005
Solutions to Problem Set #1 1-1 pg. 12 #9 Bn = i=n Ai , Cn = i=n Ai a) Bn Bn+1 . Bn = An ( i=n+1 Ai ) = An Bn+1 s Bn+1 s Bn+1 An = Bn Cn Cn+1 . Cn = An Cn+1 s Cn = An Cn+1 s Cn+1 b) n=1 Bn s Bn for all n s s i=1 Ai for al
18.05 Lecture 8 February 22, 2005
3.1 - Random Variables and Distributions Transforms the outcome of an experiment into a number.
Definitions:
Probability Space: (S, A, P)
S - sample space, A - events, P - probability
Random variable is a function on