Example 7. A poker hand, 5 cards dealt at random from a deck of 52 cards (4 suits 13
P(exactly 2 A in 5-card poker hand) =?.
Solution 1. In class, we said, we can think the sample space S as all possible combinations
of 52 cards tak
Discrete probability models
A model is a simplied representation of some aspect of the reality.
Why simple probability models?
Capture common features in commonly seen random experiments.
Summarize probabilities dened on a complex sample space using jus
Denition 1 (Gamma function). The gamma function is dened as
y 1 ey dy.
Properties of the gamma function:
1. (1) = 1.
2. () = ( 1)( 1).
3. (n) = (n 1)!, for integers n = 1, 2, 3, . . . .
4. (1/2) = .
For property 2, use the integratio
Random variables and their distributions
Read textbook 2.11, 3.1 - 3.3
A random variable assigns a number to each sample point in the sample space (assigns a
number to each outcome of a random experiment). Formally, it is a function from
Read textbook 4.1 - 4.4.
It is time to review your calculus!
Continuous random variables
Random experiments can have continuous outcomes:
X : body height of randomly selected person in a population.
Y : the waiting time at a bus stop.
Z : the time to f
Laws (rules) of probability
(Read textbook 2.8, 2.9)
The Additive Law
P(A B ) = P(A) + P(B ) P(A B ).
For three events,
P(A B C ) = P(A) + P(B ) + P(C ) P(A B ) P(A C ) P(B C ) + P(A B C ).
Only when A and B are mutually exclusive or disjoint, do we have
Conditional probability and independence
(Read textbook 2.7)
Conditional probability of an event A given that an event B (P(B ) > 0) has occurred:
P(A|B ) =
P( A B )
P( B )
Remark 1. When computing a conditional probability, we v
Read textbook 2.10.
The Law of Total Probability
The law of total probability
P(A) = P(A B ) + P(A B ).
In terms of conditional probabilities,
P(A) = P(A|B ) P(B ) + P(A|B ) P(B ).
Generally, suppose the sample space is partitioned into B1 ,B2 , . . . , B
The product rule of counting
The product rule (mn rule Page 41): the experiment takes k steps, the number of possible
outcomes in each step is ni , then the total number of outcomes is n=1 ni .
Example 1. Toss a coin 3 times, tot
Read textbook 4.5.
We can dene a continuous distribution by specifying its probability density function (p.d.f.).
Denition 1. A random variable Y is said to have a normal distribution with parameters
, 2 , such that < < and > 0, if and