More Probability

• The probablilty of an event E is given by

            Total number of ways E can occur
     P(E) = --------------------------------
           Total number of possible outcomes
     

• Example: Problem 2, page 591

  A.  We care about the order, so this is a  permutation
      So there are 6! different ways they can line up.
      or 6P6 = 6!/(6-6)! = 6!
  
  B.  3×3×2×2×1×1 = 3!×3!
  C.    (3!×3!)/6! = 1/20 
     

• Example 4

     Note, these are permutations.
     Total of 7! different orders.
  
  A. P(D appears first) 
      number of ways D can occur first 1!*6!
   
      P(D) = 6!/7! = 1/7
  
  B. C will perform 5th and B performs last
      n(E) = 5 * 4 * 3 * 2*  1 * 1* 1  = 5!
  
      P(E) = 5!/7! = 1/7*6
  
  C. 1/7!
  D. n(E) = 2!×6!, P(E) = 2!/7 = 2/7
     

• Number 6

      Note these are combinations.
  
      n(s) = 11C4
          = 11!/(4!*7!)
  	= 11*10*9*8/4*2*3 
  	= 11*10*3 
  	= 330
  
      n(E) = 6C4 
           = 6!/4!*2!
  	 = 6*5/2 = 15
   
      P(E) = 15/330 
           = 1/22
     

• Problems 8

  Notice, order of the numbers does not matter.
  
  n(S) = 30C5
       = 30!/(25)!*5!
       = 30×29×28×27×26/5×4×3×2
       = 29×14×27×13
       = 142,506
  
  n(E) = 1
  P(E)  = 1/142,506
     

• Number 10, combination or permutation
• Number 12, Combination or permutation