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You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), and K (king) of that suit. S = spades, H = Hearts, D = Diamonds, C = Clubs.

  1. Suppose you pick four cards, but do not put any cards back into the deck. Your cards are QS , 1 D , 1 C , QD .
  2. Suppose you pick four cards and put each card back before you pick the next card. Your cards are KH , 7 D , 6 D , KH .

Which of a. or b. did you sample with replacement and which did you sample without replacement?

a. Without replacement; b. With replacement

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You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), and K (king) of that suit. S = spades, H = Hearts, D = Diamonds, C = Clubs. Suppose that you sample four cards without replacement. Which of the following outcomes are possible? Answer the same question for sampling with replacement.

  1. QS , 1 D , 1 C , QD
  2. KH , 7 D , 6 D , KH
  3. QS , 7 D , 6 D , KS

without replacement: 1. Possible; 2. Impossible, 3. Possible

with replacement: 1. Possible; 2. Possible, 3. Possible

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Mutually exclusive events

A and B are mutually exclusive events if they cannot occur at the same time. This means that A and B do not share any outcomes and P ( A AND B ) = 0.

For example, suppose the sample space S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Let A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8}, and C = {7, 9}. A AND B = {4, 5}. P ( A AND B ) = 2 10 and is not equal to zero. Therefore, A and B are not mutually exclusive. A and C do not have any numbers in common so P ( A AND C ) = 0. Therefore, A and C are mutually exclusive.

If it is not known whether A and B are mutually exclusive, assume they are not until you can show otherwise . The following examples illustrate these definitions and terms.

Flip two fair coins. (This is an experiment.)

The sample space is { HH , HT , TH , TT } where T = tails and H = heads. The outcomes are HH , HT , TH , and TT . The outcomes HT and TH are different. The HT means that the first coin showed heads and the second coin showed tails. The TH means that the first coin showed tails and the second coin showed heads.

  • Let A = the event of getting at most one tail . (At most one tail means zero or one tail.) Then A can be written as { HH , HT , TH }. The outcome HH shows zero tails. HT and TH each show one tail.
  • Let B = the event of getting all tails. B can be written as { TT }. B is the complement of A , so B = A′ . Also, P ( A ) + P ( B ) = P ( A ) + P ( A′ ) = 1.
  • The probabilities for A and for B are P ( A ) = 3 4 and P ( B ) = 1 4 .
  • Let C = the event of getting all heads. C = { HH }. Since B = { TT }, P ( B AND C ) = 0. B and C are mutually exclusive. ( B and C have no members in common because you cannot have all tails and all heads at the same time.)
  • Let D = event of getting more than one tail. D = { TT }. P ( D ) = 1 4
  • Let E = event of getting a head on the first roll. (This implies you can get either a head or tail on the second roll.) E = { HT , HH }. P ( E ) = 2 4
  • Find the probability of getting at least one (one or two) tail in two flips. Let F = event of getting at least one tail in two flips. F = { HT , TH , TT }. P ( F ) = 3 4
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Source:  OpenStax, Introductory statistics. OpenStax CNX. May 06, 2016 Download for free at http://legacy.cnx.org/content/col11562/1.18
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