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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
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2.2 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 Xx1 x2…xK … Pk p1 p2 … pk …
2. Characteristics of distribution law (1) pk 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
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 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 XB (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
ExampleA 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{X2}=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
Now, lets try to solve the aforementionedproblem by putting =np=(400)(0.02)=8, then approximatelywe have P{X2}=1- P{X=0}-P {X=1} =1-(1+8)e-8=0.996981. Poisson distribution-P32 X~P{X=k}= , k=0, 1, 2, … (0)
Poisson distribution-P32 Example 2.10, 2.11 –P32
Discrete r.v. Random variable Several important r.v.s Distribution law 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