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There exists a company that makes wooden boxes. A machine used in the creation of these boxes typically averages 2 dots on each box. What is the probability that the total number of dots on 5 randomly selected boxes is at most 3.

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    One must know the *distribution* of dots---not merely the average. For instance, it could be that there are *always two dots* on each box. Alternatively, there could always be 200 dots on every 100th box. Or 20000 dots on every 10000th box. The answer to your problem depends on which of these (or other) distributions are at play. – 2017-02-28
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    @DavidG.Stork The question I am basing my question on does not specify the distribution. The information I have given is all that is in the question. – 2017-02-28
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    What probability distributions have you recently studied? Whomever set this homework problem may have assumed you would model this process on one of those. – 2017-02-28
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    Poisson and Binomial – 2017-02-28
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    @Ibrahim Abouhashish: The Poisson Distribution is the appropriate one here. (Can you see why?) – 2017-02-28
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    Well I understand its use for if I was looking at 1 box and the number of dots on that one box. But for 5 boxes how would I proceed – 2017-02-28
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    Shorthand approach: If each of the five boxes is independent and Poisson with rate $2$, then the five boxes together are also Poisson, and with rate $5 \times 2 = 10$. – 2017-02-28
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    OP: Please note that the answer you accepted is currently (as of this writing) incomplete. One additional case needs to be considered. Please see my comment under his answer. ETA: It's now been fixed. – 2017-02-28

2 Answers 2

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The cases in question are:

  • $0$s on all 5 boxes
  • $0$s on four boxes, a $1$ on one box
  • $0$s on four boxes, a $2$ on one box
  • $0$s on four boxes, a $3$ on one box
  • $0$s on three boxes, a $1$ on each of two boxes
  • $0$s on two boxes, a $1$ on each of three boxes
  • $0$s on three boxes, a $1$ on one box a $2$ on another box

Use the binomial expansion to find how many ways you can have each of these cases. For instance, there is only ${5 \choose 0} = 1$ way to have $0$s on all boxes; there are ${5 \choose 1} = 5$ ways to have $0$s on four boxes and a $1$ on another box; and so on. Then use your Poisson Distribution (with $\mu = 2$) to calculate the probability for each of the conditions. For instance, in the condition of a $1$ on exactly one box, the contribution to the probability would be: $5 P[\mu=2,0]^4 P[\mu=2,1]$.

And add them up!

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    For "0s on three boxes, a 1 on one box a 2 on another box" What would the binomial expansion be? – 2017-02-28
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    $20\ P[2,0]^3\ P[2,1]\ P[2,2]$. 20 because 5 candidate boxes for the $2$, leaving $4$ remaining boxes for the $1$ (and the $0$s must go in the remaining three boxes). – 2017-02-28
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    Shouldn't it be 15, with $\binom 5 2$, $\binom 4 1$ and $1$? – 2017-02-28
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    @IbrahimAbouhashish: No. You're confusing the *value* of the box dots with the *number of boxes where they might appear*. There are five boxes where you can put the single $2$. That leaves four boxes where you can put the single $1$. The remaining three $0$s are forced into the remaining boxes: $5 \times 4 = 20$. – 2017-02-28
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Another approach: If the number of spots on each box is independent and Poisson with rate $2$, then the number of spots on five boxes together is Poisson with rate $5 \times 2 = 10$.

You can then just add up

\begin{align} P(\text{spots} \leq 3) & = \sum_{k=0}^3 \frac{10^k}{k!} e^{-10} \\ & = \left( 1 + 10 + 50 + \frac{500}{3} \right) e^{-10} \\ & = \frac{683}{3} e^{-10} \doteq 0.01034 \end{align}

Some comments on the conditions under which you Poisson distributions can be combined or decomposed into other Poisson distributions is given at the Wikipedia entry for Poisson distribution. Independence (which we assume here, possibly recklessly) is a central factor.