45-733 PROBABILITY AND STATISTICS I Notes #5A


February 2000



Binomial Distribution

  1. Let X1, X2, X3, ..., Xn be random variables corresponding to n Bernoulli trials (experiments), and let Z be the random variable:
    Z = åi=1,n Xi
    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 + ... + Xn] = E(X1) + E(X2) + ... + E(Xn) =
    p + p + ... + p = np
    VAR(Z) = VAR[X1 + X2 + ... + Xn] = VAR(X1) + VAR(X2) + ... + VAR(Xn) =
    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