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Independent and dependent events-conditional probability

  1. Feb 4, 2012 #1
    Lets say I roll 2 dice.

    We have 36 possible elementary events.

    I want to know what is the probability that I rolled an even number, given that I rolled both same dice.

    So my event A={<1,1>, <2,2>, <3,3>, <4,4>, <5,5>, <6,6>}
    My event B={<2,2>, <4,4>,<6,6>}

    Conditional probability is,

    P(B|A)=P(A and B)/P(A) = 0.5

    I mean its intuitive, if I rolled the both dice with the same number I have 50:50 percent chance that I got an even number.

    Now are these events independent? My gut is telling me that they aren't and math is confirming that.

    This is an example I worked out myself, to try to explain this to myself, so I just need somebody to confirm it.
  2. jcsd
  3. Feb 4, 2012 #2

    Stephen Tashi

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    You haven't done the math to test whether the events are independent.
    Is P(A|B) = P(A)? Is P(B|A) = P(B)?
  4. Feb 4, 2012 #3


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    Hi Bassalisk! :smile:
    hmm :rolleyes: … let's restate the question, to make it easier :wink:

    "what is the probability that the second die has the same parity (remainder on division by 2) as the first die?"​
  5. Feb 4, 2012 #4
    It doesn't give those equalities, so they are dependent?
  6. Feb 4, 2012 #5
    Hmmm I don't think I follow. I am as good with probability, as our politicians are with our countries.
    Last edited: Feb 4, 2012
  7. Feb 4, 2012 #6

    Stephen Tashi

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    What doesn't give those inequaliies? You haven't shown any computation for P(B) or P(A).

    By the way, your notation B = {(2,2),(4,4),(6,6) } is incorrect. That set should be named "B|A".

    The set "B" is {(2,1),(2,2),(2,3)...(4,1),(4,2)... etc.
  8. Feb 4, 2012 #7
    Give me a minute.

    I thought B was my second event.

    I wanted to both dice have the same and even number. Those were 22 44 66.

    Probability that I rolled same numbers on both dice is: 6/36 right?

    Probability that I rolled same numbers on both dice, and they are even, is 3/36


    P(A and B)=3/36 (intersection between my original 2 sets)

    P(B|A)=P(A and B)/P(A)

    That is

    P(B|A)=3/36/6/36 =1/2

    If I carry on that test we spoke of:

    P(A and B)=P(A)*P(B)

    P(A and B)=6/36 * 3/36 = 18/36 which is not P(A and B) which equals 3/36

    Where am I making a mistake, why did you define that second set the way you did?
  9. Feb 4, 2012 #8

    Stephen Tashi

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    There is an important convention involving the notation P(X|Y).

    In the notation P(X), X is an "event" or set in some space of events. The space in your in your problem is all possible outcomes of rolling two fair dice = {(1,1),(1,2)..(1,6)...(6,1)...etc. }

    The notation P(X|Y) expresses the idea that you change both the set X that is considered and also the space of outcomes. The space of outcomes is now the event Y, not the entire space implied in the notation [itex] P(X) [/itex] The event that we consider within that new space of outcomes is the set [itex] X \cap Y [/itex]

    Both the notation [itex] P(X \cap Y) [/itex] and [itex] P(X|Y) [/itex] refer to the event given by [itex] X \cap Y [/itex] but [itex] P(X|Y) [/itex] refers to that set as an event in "smaller" probability space than the space implied by [itex] P(X \cap Y) [/itex].

    B is the event "you roll an even number". The standard mathematical interpretation of that is that you roll at least one even number in two rolls. This includes the cases where you roll an even number only on the first roll, the cases where you roll an even number only on the second roll and the cases where you roll an even number on both rolls. That's why I listed such possibilities.

    The event B|A refers to the event "you roll and even number within the space of events defined by both rolls being the same". This is the set {(2,2),(4,4),(6,6)} which you listed.
  10. Feb 4, 2012 #9
    Hmmm. But what is the probability P(B|A) then?. I want to find the probability, that i rolled both even numbers, given that I rolled 2 same numbers before. 50% chance is something that is intuitive to me...
  11. Feb 4, 2012 #10

    Stephen Tashi

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    I agree with your work and your answer for P(B|A).

    However, If you are writing out the answer to this problem, you need to label the sets correctly.

    You should indicate how you computed P(B) if you use that number. In your work, It looks like you are saying P(B) = 3/36. I don't think that is correct.
  12. Feb 4, 2012 #11
    I think he's getting confused.

    They are not independent. A=two dice are the same. B=two dice are even.

    P(AB) = probability of the intersection = 3/36 =1/12

    But P(A) = 1/6 and P(B) = 1/4 (You can have ee, oo, eo, oe all with equal prob)

    Since P(AB) is not equal to P(A)P(B) then the events are not independent.

    In fact, P(B given A) = P(AB)/P(A) = (1/12)/(1/6) = 1/2
  13. Feb 4, 2012 #12
    Makes sense thank you!
  14. Feb 4, 2012 #13
    Yes I am sloppy with the notation. Thank you for addressing that, I am really trying to grasp this concept.

    Again, you have been very helpful. I understand what I was missing now.
  15. Feb 4, 2012 #14
    You are welcome. Keep it simple.
  16. Feb 4, 2012 #15

    Stephen Tashi

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    OK, I see that point of view.

    The question is whether the problem says

    1. "In a throw of two fair dice, is the event of rolling an even number (meaning at least one die shows an even number) independent of the event of rolling two identical numbers?"

    Or is the problem:

    2. "In a throw of two fair dice is the event of rolling an even number given that the two dice show identical numbers independent of the event that the two dice show identical numbers."

    I was assuming the question was 1. If the equestion is 2, then my criticism of how the set B is labeled is incorrect. Bassalisk, what is the precise statement of the problem?
  17. Feb 4, 2012 #16
    That 2nd one. I was asking this:

    If I rolled 2 identical numbers. like 11 or 22 or anything from my set A. What is the chance that I got both even numbers? 2nd part of the question was:

    Are these 2 events independent of each other. That is: Is having rolled both even numbers, dependent of rolling 2 same numbers.
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