Conditional Probability

Discussion

• Consider the problem of sampling from a deck with replacement.
  • What is the probability that the second card is an ace?
  • This depends on if the first card is an ace
  • 4/51 or 3/51
• Such problems can be solved with conditional probability.
• Conditional probability is the probability that event E₂ will happen after event E₁ happens.
• It is written P(E₂|E₁)
• Example 1

  A card is selected from a deck of cards.  What is the probability that 
  it will be a 7, given that it is not a face card.
  
  If it is not a face card, that means that it must be a 1-10
  There are 40 cards 1-10
  4 of these cards are a 7
  
  P(7|not a face card)  = 4/ 40  = 1/10
     

• Example 2,

  A die is tossed.  What is the probability that the number will be 
  a 6, given that the number is even?
  
  There are 3 results that are even.
  There is 1 result that is a six.
  
  P(6|even) = 1/3
     

• We have a formula

                   n(E1 and E2)
     P(E2| E1) =  -------------
                      n(E1)
     

• Example 3 (#6, page 710)

  
     n(yellow) = 2)
     n(yellow and 3) = 0
  
     P(3|yellow) = 0/3 = 0;
     

• Example 4, #8

      n(N < 5) = 4
      n(N < 5 and N < 3) = 2
  
      P( N < 3|N < 5) = 2/4 = 1/2
     

• Some of these problems require us to construct a tree diagram.
• Example 5, 30, 32, 34

  Toss a die twice
  
  Draw a tree, sample space
  
  1-1 1-2 1-3 1-4 1-5 1-6 
  2-1 2-2 2-3 2-4 2-5 2-6
  3-1 3-2 3-3 3-4 3-5 3-6
  4-1 4-2 4-3 4-4 4-5 4-6
  5-1 5-2 5-3 5-4 5-5 5-6
  6-1 6-2 6-3 6-4 6-5 6-6
       P(sum=6|die1=1)  = 1/6
           n(die1=1) = 6
  	 n(sum-6 and die1 = 1) = 1 (1-5)
       P(sum even | die2=2) 3/6 = 1/2
           n(die2=2) = 6
  	 n(die2=2 and sum is even) = 3  (2-2, 4-2, 6-2) 
       P(7 or 11 | die1 = 5) = 2/6 = 1/3
           n(die1 = 5) = 6
  	 n(sum = 7 or 11) = 2   5-2 or 5-6
     

• Example 6, 38

        n(men) = 105
        n(men and prefers coke) = 60
        P(prefers coke | man) = 60/105
     

Homework