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Conditional Probability and the Independence of Events

  1. Jun 27, 2009 #1
    I'm having trouble with one of the rules of probability

    P(A n B) = P(A)P(B) which holds if events A and B are independent

    The following problem illustrates my confusion. I've defined Events A and B below, are these events dependent? Per the solution in the book P(A [tex]\cap[/tex] B) = P(A)P(B), that is the only way that P(A|B) will equal P(A) therefore by the solution A and B are independent

    But how are they independent if the probability of selecting the next 3 cards,

    P(A)= ([tex]\stackrel{11}{3}[/tex] ) / ([tex]\stackrel{50}{3}[/tex])

    depends on knowing there are now only 50 cards to choose from

    Moreover the solutions manual says this problem is like P(S3 S4 S5 |S1 S2) but if this is the case I see no intersection between S3 S4 S5 and S1 S2, therefore P(S3 S4 S5 [tex]\cap[/tex] S1 S2) = 0 and P(S3 S4 S5 [tex]\cap[/tex] S1 S2) != P(S3 S4 S5) P(S1 S2)

    So

    1) I'm not sure how these events are independant, which is the assumption taken to arrive at the answer and if they are dependent then we can't equate P(A intersect B)/P(B) to P(A)P(B)/P(B) in the solution

    2) I'm a bit confused about the intersection of A and B, how is the intersection not disjoint?
    Am I making a wrong assumption that A and B have nothing in common?




    HERE IS THE PROBLEM:

    Cards are dealt, one at a time, from a standard 52-card deck.
    If the first 2 cards are both spades, what is the probability that the next 3 cards are also spades?

    we break this problem down into two events
    A: remaining 3 cards are spade
    B: first 2 cards are spade


    P(A)= ([tex]\stackrel{11}{3}[/tex] ) / ([tex]\stackrel{50}{3}[/tex])

    P(B)= ([tex]\stackrel{13}{2}[/tex] ) / ( [tex]\stackrel{52}{2}[/tex] )


    P(A|B) = P(A [tex]\cap[/tex] B)/P(B)

    According to the book I'm working with, the answer is

    P(A|B) = P(A [tex]\cap[/tex] B)/P(B)= P(A)P(B)/P(B) = P(A) = .008418
     
    Last edited: Jun 27, 2009
  2. jcsd
  3. Jun 27, 2009 #2

    D H

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    Your event A is already a conditional event.

    Suppose you deal out five cards face down. Now split the first two cards dealt from the last three. Call these two sets b and c. Your event B is equivalent to both members of set b being spades. Define event C to be all three of set c being spades. Note very well: Event C is different from your event A. In fact, your event A is my event C given event B: A=C|B .

    So what is the event A|B ? It is event C given event B given event B. That second "given" doesn't change things a bit. That P(A|B)=P(A) is a given in this case. That a conditional event is statistically independent of the event upon with the event is conditioned in a sense is a tautology.
     
    Last edited: Jun 27, 2009
  4. Jun 27, 2009 #3

    EnumaElish

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    Just a cautionary note: P({S3 S4 S5} and {S1 S2}) = P({S3 S4 S5 S1 S2}), which is definitely not zero!
     
  5. Jun 28, 2009 #4

    I have a different perspective here, I don't know how P({S3 S4 S5} and {S1 S2}) = P({S3 S4 S5 S1 S2}). In regards to the intersection of two sets.


    By definition, the intersection of two sets equals the elements from each set that belong to boths sets. What elements do {S3 S4 S5} and {S1 S2} have in common? I know multiplying the probability of each individually by the other gives the right answer in this case, but what elements do {S3 S4 S5} and {S1 S2} have in common?

    - {S3 S4 S5} != {S1 S2}
    -{S1 S2} ! [tex]\subseteq[/tex] {S3 S4 S5} and vice versa

    I don't see an intersection here. How do I adjust my view to see this?
     
  6. Jun 28, 2009 #5

    statdad

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    I think EnumaElish was saying that, when considered in sequence, drawing two spades first (S1 & S1) followed by three more (S3 & S4 & S5) corresponds to the single sequence (S1 S2 S3 S4 S5), and that the probability of drawing five spades during five draws is not zero.
    The notation chosen wasn't the best, and at first I too wondered at the point.

    EnumaElish, if I misinterpreted your post I apologize.
     
  7. Jun 28, 2009 #6

    EnumaElish

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    Thanks statdad; you have the correct interpretation.

    I had not clarified that I was using the set notation to denote an event -- apologies for the confusion.

    Event 1 = 2 spades being drawn
    Event 2 = 3 spades being drawn.
    Event 1 and event 2 = 5 spades being drawn.
     
  8. Jun 29, 2009 #7
    I was playing around with the numbers where
    A = 11 C 3
    B = 13 C 2
    S = 13 C 5

    AB != S, however

    P(A) = 11 C 3/50 C 3
    P(B) = 13 C 2/ 52 C 2
    P(S) = 13 C 5/52 C 5

    (11 C 3/50 C 3)(13 C 2/ 52 C 2) == (13 C 5/52 C 5)

    P(A)P(B) == P(S)

    or the equivalent of EnumaElish's P(S3S4S5)P(S1S2) == P(S1S2S3S4S5)

    So numerically it works out. Moreover I found some documentation online stating something that is apperantly extremely obvious (since it's rarely mentioned)

    Probability that two independent events Ev(X) and Ev(Y), will happen simultaneously or sequentially:

    Pr(XnY) = Pr(X)Pr(Y)

    which is more information than what the book gives me, which is:

    Two events A and B are said to be independent if any one of the following holds:

    Two definitions involving conditional events and
    P(AnB) = P(A)P(B)


    Moreover, this problem is thrown in with a bunch of problems that have an actuall countable intersection, or have a probability value for the intersection in a table so you can actually compare given values for P(AnB) and P(A)P(B)

    Lastly one important fact is left out, correct me if i'm wrong.

    the probability of the intersection of two simultaneous independent events is the product of the probability of each event, and vice versa (a restatement of the online ref I found)

    Pr(XnY) = Pr(X)Pr(Y)


    But tell me something, I found this on wiki
    http://en.wikipedia.org/wiki/Independent_events

    "By contrast, if two cards are drawn without replacement from a deck of cards, the event of drawing a red card on the first trial and that of drawing a red card on the second trial are dependent."

    However, in my problem, where cards (Spades) are also drawn without replacement, A and B are independent but in the event just cited, the trials are dependent. Can someone shed some light on the differences here. Why the events i'm dealing with are independent, but the ones above are dependent, when they seem to share similar characteristics?
     
    Last edited: Jun 29, 2009
  9. Jun 29, 2009 #8

    D H

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    Your event A is the probability of drawing three spades from a deck from which two spades have been removed prior to drawing three cards. Suppose you had instead defined event A to be the probability of drawing three spades from a full deck. Call this A' to distinguish it from A. If you go through the math you will find that P(S) is not equal to P(B)*P(A'). A' and B are not independent events. These events B and A' are what the wikipedia article was alluding to, not your events B and A.

    A conditional event is by definition statistically independent from the event upon which it is conditioned. It is a tautology and doesn't have much meaning beyond that.
     
  10. Jun 29, 2009 #9
    DH, thanks for the clarification on the wiki, and thanks for hammering in the fact that

    "A conditional event is by definition statistically independent from the event upon which it is conditioned."

    I think it's finally sinking in. Perhaps my recent denseness is due to a loss of focus in my independent summer study.
     
  11. Jun 29, 2009 #10
    I just want to clarify, for my sake, the difference in the wiki citation and my problem, correct me if i'm wrong.

    "By contrast, if two cards are drawn without replacement from a deck of cards, the event of drawing a red card on the first trial and that of drawing a red card on the second trial are dependent."

    A: Drawing a red card second trial
    B: Drawing a red card first trial

    b/c the description of event A is not conditioned on B, the events are dependent.

    Ok, I think i'm starting to see the big picture here.
    There are many ways to describe events, (two events in this case) somethings to watch out for:

    Event coming from same population (deck of Cards) where they describe two samplings
    - with replacement (2 events indep)
    - without replacement
    i. description of second event not conditioned on first (2 events dep)
    ii. description of second event conditioned on first (2 events indep)

    correct me if i'm wrong.
     
  12. Jun 29, 2009 #11

    EnumaElish

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    I don't see the point of you referring to my PF userid as "enumafish." I find it slightly irritating.

    It's not as if I'm getting paid to help you out here; AFAIK most if not all people are not getting paid for this. The least you can do is to try to keep that in mind, thank you.
     
  13. Jun 29, 2009 #12
    EnumaElish,

    The only point it may have is to illustrate my absentmindedness. Apologies for my carelessness, in no way did I intend disrespect towards you now and in any of our previous interaction, I mean that with all sincerity. I'm sure I would be perturbed as well if someone referred to me by a wrong name, thank you for bringing this to my attention. I have changed the post to reflect your correct userID. I am grateful for your insight and all others in the PF community.
     
  14. Jun 29, 2009 #13

    jim mcnamara

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    Enuma Elish is the anglicized spelling (Enuma Eliš is another) of the incipit from the Babylonian creation story.

    See: W. C. Lambert, S. B. Parker, Enuma Eliš. "The Babylonian Epic of Creation", Oxford 1966.

    FWIW.
     
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