Conditional Probability

• If a card is drawn from a deck of cards, determine the probability that it is a 6, given that it is a number card.
• P(E₂ | E₁) = n(E₁ and E₂)/n(E₁)
• P(6 | number card) is the probability of a 6, given that it is a number card.
• There are 4 cards which are both 6 and a number card
• There are 40 number cards
• P(6 | number card) =4/40 or 1/10
• Try numbers 17, 19, 29-34, 37

The Counting Principle and Permutations

• The Counting Principle states that if one experiment has M outcomes and a second experiment has N outcomes then the two experiments together have M × N outcomes.
• A Permutation is an ordered arrangement of objects.
• n! = n × (n-1) × (n-2) ... 3 × 2 × 1
• 5! = 5 × 4 × 3 × 2 × 1 = 120
• 0! = 1 (by definition)
• If I have 4 dice, how many ways can I display them? 4! = 24
• If I have 5 dice, how many ways can I display them? 5! or 120
• If I have 5 dice, but can only display 2, how many ways can this be done?
• This is a _nPr problem.
• In this case, n = 5 and r = 3
• _nPr = n!/(n-r)!
• ₅P3 = 5!/(5-3)! = 5!/2! = 60
• do 9-20
• do 22, 23, 31,
• ORDER IS IMPORTANT FOR THESE QUESTIONS.
• If I have duplicate objects (like letters in a word)
• the number of permutations is n!/[n₁!×n₂!×...n_k!]
• try 56

      mississippi
      m =1
      i = 4                   11!
      s = 4               -----------
      p = 2               1!×2!×4!×4!