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§2. Discrete random variable and its distribution law
1.Discrete random variables
Definition 2.2 –P27
Definition 2.3 –P27
Definition 2.4 Suppose that r.v. X assume value x1, x2,
…, xn, … with probability p1, p2, …, pn, …respectively, then it
is said that r.v. X is a discrete r.v. and name
P{X=xk}=pk, (k=1, 2, … )
the distribution law of X.
The distribution law of X can be represented by
X～ P{X=xk}=pk, (k=1, 2, … )，
or
X
x1
x2
…
xK
…
Pk
p1
p2
…
pk
…
X~
2. Characteristics of distribution law
(1) pk  0, k＝1, 2, … ;
(2)
 p ＝1.
k 1
k
Example 1 Suppose that there are 5 balls in a bag, 2 of them
are white and the others are black, now pick 3 ball from the
bag without putting back, try to determine the distribution
law of r.v. X, where X is the number of white ball among the 3
picked ball. In fact, X assumes value 0，1，2 and
C2k C33 k
P{ X＝k}＝
.
3
C5
Example 2.2
P27
Several Important Discrete R.V.
(0-1) distribution
let X denote the number that event A appeared in a trail,
then X has the following distribution law
X～P{X＝k}＝pk(1－p)1－k, (0<p<1) k＝0，1
or
X
1
pk
p
0
1 p
and X is said to follow a (0-1) distribution
Binomial distribution
Let X denote the numbers that event A appeared in a nrepeated Bernoulli experiment, then X is said to follow a
binomial distribution with parameters n, p and represent it
by XB (n, p). The distribution law of X is given as :
P{ X  k}  C n p (1  p)
k
k
nk
, (k  0,1,..., n)
Bernoulli Trial
-P29
one of a sequence of independent
experiments each of which has the same
probability of success
1. n-repeated
2. two possible outcomes :success & failure
3. P (success)=p
4. independent
Example 2.5, 2.6, 2.7
P29-31
Example A soldier try to shot a bomber with probability
0.02 that he can hit the target, suppose the he
independently give the target 400 shots, try to
determine the probability that he hit the target at least
for twice.
Answer Let X represent the number that hit the target in 400
shots, Then X～B(400, 0.02), thus
P{X2}＝1－ P{X＝0}－P {X＝1}＝1－0.98400-(400)(0.02)(0.98)399
=…
Poisson theorem If Xn~B(n, p), (n＝0, 1, 2,…) and n is large
enough, p is very small, denote =np，then
P{X  k} 
k
k!
e  ,
k  0,1,2,...
Now, lets try to solve the aforementionedproblem by
putting  =np＝(400)(0.02)＝8, then approximatelywe
have
P{X2}＝1－ P{X＝0}－P {X＝1}
＝1－(1＋8)e－8＝0.996981.
Poisson distribution-P32
k

X～P{X＝k}＝
e , k＝0, 1, 2, … (0)
k!
Poisson distribution-P32
Example 2.10, 2.11 –P32
Random variable
Discrete r.v.
Distribution law
Several important r.v.s
0-1 distribution
Poisson distribution
Bionomial distribution
Uniform Distribution
X
pk
a1
1
n
P28
a 2 a n
1
1

n
n
Example 2.3, 2.4 –P28
Geometric Distribution P31
Example 2.8, 2.9 –P31
P48：
Exercise 2：Q1，2，3