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I got this question in my homework but i cant figure out the procedure. i don't want an exact solution but i just want a technique or you can say hint. i just want to which specific distribution it is related binomial,poisson etc

If n ≥ 1 balls are placed randomly into M ≥ 4 boxes such that each ball has equal chance of adding in any of the M boxes and the placement of different balls is independent. Let X is the number of balls in the first box and Y be the number of balls in the first four boxes.

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    $\begingroup$ For each ball there is $1/m$ probability of going into any given box. $n/m$ gives you the expected number of balls in any box. $E(Y)=4n/m$. $X$ is easily computed plugging $1/m$ into the Binomial formula. For $Y$ you can sum for the first 4 boxes. The binomial formula you can easily find. If $p=1/m$ then $1-p$ is the probability of a ball going elsewhere. Binomial measures probability of a given number being in one box by multiplying the probability that number is in the box, by the probability the remainder were outside the box, times the number of combinations of ways that can happen $\endgroup$ – user334732 Oct 24 '17 at 17:43
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$X$ and $Y$ are both binomial variables. For example, $X$ is the number of successes in $n$ trials, where a trial is putting a ball in a box, and success is the ball being put in the first box.

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  • $\begingroup$ it means i have to calculate probability of x and multiplied by the value x takes so in this case it no of balls. so how will i calculate mean of x here? the probability it is in first box is 1/n? $\endgroup$ – Zulfiqar Tariq Oct 24 '17 at 17:43
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    $\begingroup$ @ZulfiqarTariq mean is the expected value of $X$ which is $n/m$ as given above. $\endgroup$ – user334732 Oct 24 '17 at 17:45
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    $\begingroup$ @ZulfiqarTariq There is a formula for the mean of a binomial random variable. You just have to work out the parameters $p$ and $n$ (and understand why $X$ and $Y$ are binomial). $\endgroup$ – Jack M Oct 24 '17 at 17:50

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