45-733 PROBABILITY AND STATISTICS I Notes #5A

February 2000

Binomial Distribution

1. Let X₁, X₂, X₃, ..., X_n be random variables corresponding to n Bernoulli trials (experiments), and let Z be the random variable:
   Z = å_i=1,n X_i
   The distribution of Z is the binomial
   

   
                            æ ænöpz(1-p)(n-z)
                            ç ç ÷
                            ç èzø
                     f(z) = ç            z=0,1,2,3,...,n
                            ç
                            è 0 otherwise
   

   
   Where: E(Z) = E[X₁ + X₂ + ... + X_n] = E(X₁) + E(X₂) + ... + E(X_n) =
   p + p + ... + p = np
   VAR(Z) = VAR[X₁ + X₂ + ... + X_n] = VAR(X₁) + VAR(X₂) + ... + VAR(X_n) =
   p(1 - p) + p(1 - p) + ... + p(1 - p) = np(1 - p)
   
   
2. Problem 3.26 p.96-97
   From the problem statement: p = .8, and n = 20
   
   1. What is the probability that exactly 14 survive? That is:
      P(Z = 14) = F(14) - F(13) = .196 - .087 = .109 , Using the Table on p.722
      
   2. What is the probability that at least 10 survive? That is:
      P(Z ³ 10) = 1 - F(9) = 1 - .001 = .999 , Using the Table on p.722
      
   3. What is the probability that at least 14 but not more than 18 survive?
      P(14 £ Z £ 18) = F(18) - F(13) = .931 - .087 = .844
      
   4. At most 16 survive?
      P(Z £ 16) = F(16) = .589
      
      
3. Problem 3.34 p.96
   From the problem statement: p = .9, and n = 5
   
   1. P(Z = 4) = F(4) - F(3) = .410 - .081 = .329
      P(Z ³ 1) = 1 - F(0) = 1 - .1⁵ = 1 - .00001 = .99999
   
   
   2. P(Detect) = .999 = 1 - P(not Detect)
      P(not Detect) = .001 = (.1)ⁿ
      Hence, n ³ 3
   
   
4. Circuit Problem
   From the problem statement: p = .2, and n = 20 and Failures are Independent
   Let Z = number of components that have failed. Hence:
   P(Z ³ 2|Z ³ 1) = P(Z ³ 2 Ç Z ³ 1)/ P(Z ³ 1) = P(Z ³ 2)/ P(Z ³ 1) =
   [1 - F(1)]/[1 - F(0)] = [1 - .069]/[1 - .012] = .931/.988 = .942