2.2 Discrete Random Variables

2.2 Discrete random variables Definition 2.2 –P27 Definition 2.3 –P27

Definition 2.4 Suppose that r.v. X assume value x 1, x 2, …, x n, … with probability p 1, p 2, …, p n, …respectively, then it is said that r.v. X is a discrete r.v. and name P{X=x k }=p k, (k=1, 2, … ) the distribution law of X. The distribution law of X can be represented by X ～ P{X=x k }=p k, (k=1, 2, … ) ， or Xx 1 x 2 … x K … P k p 1 p 2 … p k …

(1) p k  0, k ＝ 1, 2, … ; (2) 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 2. Characteristics of distribution law

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} ＝ p k (1 － p) 1 － k, (0<p<1) k ＝ 0 ， 1 or and X is said to follow a (0-1) distribution

Let X denote the numbers that event A appeared in a n- repeated 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 : Binomial distribution

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. Poisson theorem If X n ~B(n, p), (n ＝ 0, 1, 2,…) and n is large enough, p is very small, denote =np ， then 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 － (400)(0.02)(0.98) 399 =…

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 ＝ Poisson distribution -P32 X ～ P{X ＝ k} ＝, k ＝ 0, 1, 2, … (  0)

Poisson distribution-P32 Example 2.10, 2.11 –P32

Random variable Discrete r.v. Distribution law Several important r.v.s 0-1 distribution Bionomial distribution Poisson distribution

Uniform Distribution P28 Example 2.3, 2.4 –P28

Geometric Distribution P31 Example 2.8, 2.9 –P31

P48 ： Exercise 2 ： Q1 ， 2 ， 3