Probability of random dots on a number of boxes.

Question

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.

Answer 1

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!

Answer 2

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}

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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.