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Show that two events are independent

  1. Sep 14, 2015 #1
    1. The problem statement, all variables and given/known data

    Two cards are drawn in succession without replacement from a standard deck. Show that the events A = ”face card on first draw” and B = ”heart on second draw” are independent.

    Hint: Write A = A1 ∪ A2, where A1 = ”face card and a heart on first draw” and A2 = ”face card and not a heart on first draw.”

    2. Relevant equations

    Two events A and B are independent if P(A ∩ B) = P(A)P(B), law of total probability, conditional probability, Bayes' Law, etc.

    3. The attempt at a solution

    The unconditional probability of A is P(A) = 12/52. A = A1 ∪ A2, A1 ∩ A2 = ∅, and P(A1 ∪ A2) = P(A1) + P(A2).

    For this case, I considered A as the "new" sample space and used the law of total probability.

    P(B | A) = P(B | A1)P(A1) + P(B | A2)P(A2).

    P(B | A) = (12/51)(3/52) + (13/51)(9/52) = 3/52.

    P(B ∩ A) = P(B | A)P(A) = (3/52)(12/52) = 36/522.

    Assuming this is correct so far - I just need to find P(B) to determine if they're independent.

    So, I let A and ~A be disjoint sets and calculate P(B) = P(B | A)P(A) + P(B | ~A)P(~A) with

    A = A1 ∪ A2 and ~A = ~A1 ∪ ~A2.

    P(B) = (3/52) + (10/52) = (13/52).

    P(A)P(B) = (12/52)(13/52) ≠ 36/522.
     
  2. jcsd
  3. Sep 14, 2015 #2

    andrewkirk

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    It went off track here.
    The first two lines are P(B ∩ A), not P(B | A).
     
  4. Sep 14, 2015 #3

    Ray Vickson

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    [tex] P(B|A) = P(B|A_1) P(A_1) + P(B|A_2) P(A_2) \; \Longleftarrow \text{false!} [/tex]
    You need to go back to first principles:
    [tex] \begin{array}{l}
    P(B|A) = \displaystyle \frac{P(B \cap A)}{P(A)} = \frac{P(B \cap A_1) + P(B \cap A_2)}{P(A)} \\
    = \displaystyle \frac{P(B|A_1) P(A_1) + P(B|A_2) P(A_2) }{P(A_1) + P(A_2)}
    \end{array} [/tex]
     
  5. Sep 14, 2015 #4
    What's wrong with my reasoning? Given that A is true, I made it the "new" sample space and used total probability for that subset.
     
  6. Sep 14, 2015 #5

    andrewkirk

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    I think Ray's post explains what's wrong quite nicely.
     
  7. Sep 14, 2015 #6
    Then, P(B ∩ A) = P(B | A)P(A) = (1/4)(12/52) = 3/52.

    P(A) = 12/52
    P(B) = 13/52

    P(A)P(B) = 156/2704 = 3/52
     
    Last edited: Sep 15, 2015
  8. Sep 15, 2015 #7

    Ray Vickson

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    OK, now. However, I found your explanation of P(B) = 13/52 unconvincing; it is almost as though you arrive at the correct answer by accident, using incorrect reasons. You say that if ##A = A_1 \cup A_2## then ##\sim A = (\sim A_1) \cup (\sim A_2)##, and that is false (just look at a Venn diagram to see why). There is a well-known correct expression for ##\sim(A_1 \cup A_2)## which you surely must have seen before. I really could not follow your steps; perhaps you did correct calculations but labelled them incorrectly.

    Just for the record: an easy way to get P(B) = 13/52 in cases like this one is to note that the unconditional probability P{heart on draw n} = 13/52 for any n = 1,2,3,...,52. Yes---even on draw 52, a draw with only one card left! The reason is that if you regard the sample space as the set of all 52! permutations of the numbers 1, 2,..., 52, then P{heart on draw n} = N_n/52!, where N_n = number of permutations where a 'heart' occupies slot n.This is obviously (and provably) the same whether n = 1, or n = 2, ..., or n = 52.
     
  9. Sep 15, 2015 #8
    *It was unconvincing to me, too. :wink:

    Well, it should be ~(A1 ∪ A2) = (~A1 ∩ ~A2). However, I considered ~A as non-face-card/heart and non-face-card/non-heart. I realize now that the notation was confusing. I should have made ~A = E ∪ F, or something.

    I was also considering this experiment as a 2-tuple: (first card, second card). The possible number of permutations is 52*51. Then, P(B) = P(#(whatever card, heart))/(52*51).
     
    Last edited: Sep 15, 2015
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