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Conditional Probability

Ramon distributes 12 cards into the four piles below:

Pile A

King of Clubs King of Spades 7 of Clubs
Pile B

10 of Diamonds 7 of Spades Ace of Hearts King of Hearts
Pile C

King of Diamonds 5 of Clubs Jack of Clubs 2 of Hearts
Pile D

2 of Diamonds 2 of Clubs Queen of Diamonds 8 of Clubs 7 of Hearts

Example What is the probability that...
  1. a card selected at random is a 7?
  2. a card selected at random is from pile B?
  3. a card selected from Pile D is a diamond?
  4. a card selected from above that is a King was in Pile A or Pile C?

  1. The probability a card selected at random is a 7 is P(7) = 3 / 12

  2. There are 12 possible cards that may be selected; of these, 4 are in Pile B. Therefore P(card from Pile B) = 4 / 12
  3. .

  4. There are 5 cards in Pile D. Of these 5 cards, there are 2 diamonds. Therefore, the probability that a card is a diamond given that it is in Pile D is 2 / 5

  5. This is an example of conditional probability. For any events (we'll call them A and E), the probability that A will occur given that event E has happened is P(A | E). (This is read "the probability of A, given E.") How do you find P(A | E)?

    P(A | E) = # of ways both A and E can occur = P(A and E)
    # of ways E can occur P(E)
    For this example, we are looking for the probability of getting a diamond given that we have selected a card in Pile D - this is P(diamond | card in Pile D).

    P(diamond | card in Pile E) = P(diamond and in pile D) = 2 / 12 = 2 / 5
    ________________ ____
    P(in Pile D) 5 / 12
  6. To find the probability that a randomly selected King came from Pile A or C we must find...

    P(card is in A or C | King) = # of Kings in A or in C = (# Kings in A) + (# Kings in C) = 3 / 4
    ________________ ________________
    # of Kings total 4

Think that you have perfected probability? Try the Probability Practice Problems

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