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Math Help - Expectation

  1. #1
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    Expectation

    1) A building has 10 floors above the basement. If 12 people get into an elevator at the basement, and each chooses a floor at random to get out, independently of the others, at how many floors do you expect the elevator to make a stop to let out one or more of these 12 people?

    2) Let A1, A2, and A3 be events with probabilities 1/5, 1/4, and 1/3, respectively. Let N be the number of these events that occur.
    a) Write down a formula for N in terms of indicators.
    b) Find E(N)
    In each of the following cases, calculate Var(N):
    c) A1, A2, A3 are disjoint;
    d)they are independent;
    e) A1 is a subset of A2 is subset of A3.

    I'm having a hard time figuring these out. Id really appreciacte any help!
    Last edited by tbl9301; October 19th 2009 at 05:55 PM.
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  2. #2
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    Quote Originally Posted by tbl9301 View Post
    1) A building has 10 floors above the basement. If 12 people get into an elevator at the basement, and each chooses a floor at random to get out, independently of the others, at how many floors do you expect the elevator to make a stop to let out one or more of these 12 people?

    2) Let A1, A2, and A3 be events with probabilities 1/5, 1/4, and 1/3, respectively. Let N be the number of these events that occur.
    a) Write down a formula for N in terms of indicators.
    b) Find E(N)
    In each of the following cases, calculate Var(N):
    c) A1, A2, A3 are disjoint;
    d)they are independent;
    e) A1 is a subset of A2 is subset of A3.

    I'm having a hard time figuring these out. Id really appreciacte any help!
    1) Let X_i = 1 if someone gets off on the ith floor, 0 otherwise.

    The probability that any one person gets off on floor i is 1/10, and the probability that he does not get off is 9/10. So the probability that no one gets off on floor i is (9/10)^{12}. Then

    E(X_i) = \Pr(X_i = 1) = 1 - \Pr(X_i = 0) = 1 - (9/10)^{12}.

    The expected value of the number of floors where the elevator stops is then
    E(\sum_{i=1}^{12} X_i) = \sum_{i=1}^{12} E(X_i) = \sum_{i=1}^{12} [1 - (9/10)^{12}] = 12 \cdot  [1 - (9/10)^{12}].

    Here we have made use of the theorem that E(X+Y) = E(X) + E(Y). It's important to realize that the theorem does not require that X and Y be independent. That's good for us in this problem, because the X_i's are not independent.
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  3. #3
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    Correction

    Quote Originally Posted by awkward View Post
    1) Let X_i = 1 if someone gets off on the ith floor, 0 otherwise.

    The probability that any one person gets off on floor i is 1/10, and the probability that he does not get off is 9/10. So the probability that no one gets off on floor i is (9/10)^{12}. Then

    E(X_i) = \Pr(X_i = 1) = 1 - \Pr(X_i = 0) = 1 - (9/10)^{12}.

    The expected value of the number of floors where the elevator stops is then
    E(\sum_{i=1}^{12} X_i) = \sum_{i=1}^{12} E(X_i) = \sum_{i=1}^{12} [1 - (9/10)^{12}] = 12 \cdot  [1 - (9/10)^{12}].

    Here we have made use of the theorem that E(X+Y) = E(X) + E(Y). It's important to realize that the theorem does not require that X and Y be independent. That's good for us in this problem, because the X_i's are not independent.
    Oops!

    I can't seem to keep my floors and people straight. There are 10 floors, so the summations above should run from 1 to 10 instead of 1 to 12, resulting in a a final answer of

    10 \cdot  [1 - (9/10)^{12}]
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