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Thread: Probability and expected value! Urgent

  1. #1
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    Probability and expected value! Urgent

    There are 11 boxes and 7 balls. The balls are dropped independently and uniformly at random in to the boxes. Find the expected number of boxes which contain no more than 1 ball.

    The key part of this problem is finding the probabilty of boxes that containes no more than 1 ball. But I dont know how to get there...

    Thanks very much for the help.
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    Quote Originally Posted by Betty Jan View Post
    There are 11 boxes and 7 balls. The balls are dropped independently and uniformly at random in to the boxes. Find the expected number of boxes which contain no more than 1 ball.

    The key part of this problem is finding the probabilty of boxes that containes no more than 1 ball. But I dont know how to get there...

    Thanks very much for the help.
    Hi Betty Jan,

    Let $\displaystyle X_i$ be the number of balls in box number $\displaystyle i$, where $\displaystyle 1 \leq i \leq 11$.

    Then $\displaystyle X_i$ has a Binomial distribution with $\displaystyle n = 7 \text{ and } p = 1/11$, so
    $\displaystyle P(X_i \leq 1) = P(X_i=0) + P(X_i=1) = (10/11)^7 + \binom{7}{1} (1/11) (10/11)^6 \qquad(*)$.

    Now define
    $\displaystyle Y_i = 1 \text{ if } X_i \leq 1$,
    $\displaystyle \quad = 0 $ otherwise, for $\displaystyle 1 \leq i \leq 11$,
    so $\displaystyle E[Y_i] = P(X_i \leq 1)$.

    Then the expected number of boxes with no more than one ball in each box is
    $\displaystyle E(Y_1 + Y_2 + \dots + Y_{11}) = E(Y_1) + E(Y_2) + \dots + E(Y_{11}) \qquad(**)$
    $\displaystyle = 11 \; P(X_i \leq 1)$

    where $\displaystyle P(X_i \leq 1)$ is given by (*) above.

    The theorem that E(X+Y) = E(X) + E(Y) plays a crucial role at equation (**).
    It's important to know that this theorem is true even if X and Y are not independent, which is good for us here because the $\displaystyle Y_i$'s are not independent.
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