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Homework Help: Probability that 3 people are born on different months from each other

  1. Nov 26, 2011 #1
    1. The problem statement, all variables and given/known data

    find the probability that 3 people are born on different months from each other

    3. The attempt at a solution

    So here's my reasoning. There are 3 people X1, X2, X3. Let (Xi,Xj) denote the pair have different birthmonths. P[(X1,X2) and (X2,X3) and (X3,X1)]=P(X1,X2)*P(X2,X3)*P(X3,X1)

    Well P(Xi,Xj)=11/12
    thus P[(X1,X2) and (X2,X3) and (X3,X1)]=P(X1,X2)*P(X2,X3)*P(X3,X1)=(11/12)^3

    But that is not the answer. My book said the answer is (11/12)*(10/12).

    I dont see what is wrong with my reasoning. for the 3 to have different birthmonths, any two would have different birthmonths. X1 different from X2, X2 from X3, X3 from X1.
  2. jcsd
  3. Nov 26, 2011 #2


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    Do you mean that no two are born in the same month, or that it is not the case that the three are born in the same month, and that you are assuming that the probability of being born in any month is equally-likely ?
  4. Nov 26, 2011 #3
    Sorry. The question mentions that no two people are born on the same month. Also, the probability being born in any month is equality likely.
  5. Nov 26, 2011 #4


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    Sorry, my quote function does not work. Notice that X1,X2 may have been born in different months, and same for X2,X3, but it is still possible for X1,X3 to have been born in the same month as each other.

    Basically: how many total choices of months are there for the 3 people to have been born in? And, out of all those choices, how many satisfy your condition ? One way of seeing the last is that, once X1, say has "chosen" a month to be born in, then you don't consider that choice for X2, and so on. Try a tree in which the root is the month X1 is born in, and where that month is no longer considered for X2, etc.
  6. Nov 26, 2011 #5
    I think i accounted for X1 and X3 when i set up the equation P[(X1,X2) and (X2,X3) and (X3,X1)]=P(X1,X2)*P(X2,X3)*P(X3,X1). There are only 3 combinations where any two cannot be equal to each other.
  7. Nov 26, 2011 #6
    Thanks. That makes sense. I didn't think of approaching the problem this way at first. The most intuitive way for me to approach counting isn't to take up one slot at a time but to consider the whole picture like looking at the set of 3 and then taking 3C2.
  8. Nov 26, 2011 #7


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    I agree it is a good idea to look at the big picture and to experiment, but it is also good to have a more formal way to fall back on, to check your other approaches. The issue of whether you are double-counting or not in considering (xi,xj) as you did is tricky, and so is the independence. Maybe testing against the more formal way would help.
  9. Nov 26, 2011 #8

    D H

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    This formula (the joint probability is the product of the probabilities of the individual events) is only valid if the individual events are independent of one another. What are your events, and are they truly independent events?

    BTW, the answer in your book, (11/12)*(10/12), is correct.
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