Run Probabilities for Defective Parts

probabilityrandom

This is Problems 2.110 and 2.111 in Mathematical Statistics with Applications, 5th Ed., by Wackerly, Mendenhall, and Scheaffer.

Original Question:

A machine for producing a new experimental electronic component generates defectives
from time to time in a random manner. The supervising engineer for a particular machine has noticed
that defectives seem to be grouping (hence appearing in a nonrandom manner) and thereby suggesting
a malfunction in some part of the machine. One test for nonrandomness is based on the number of
runs of defectives and nondefectives (a run is an unbroken sequence of either defectives or
nondefectives). The smaller the number of runs, the greater will be the amount of evidence indicating
nonrandomness. Of twelve components drawn from the machine, the first ten were not defective and
the last two were defective ($N\,N\,N\,N\,N\,N\,N\,N\,N\,N\,D\,D$). Assume randomness.

  1. What is the probability of observing this arrangement (resulting in two runs) given that
    ten of the twelve components are not defective?
  2. What is the probability of observing two runs?
  3. What is the probability that the number of runs $R$ is less
    than or equal to $3?$

My Answer:

  1. We have to count how many total ways we can place the $D$s in the sequence. There are basically
    $12$ places to put the first $D$, and $11$ places to put the second $D$. But the two $D$s are
    indistinguishable, so we must divide by $2$. That is, we have $132/2=66$ ways to arrange the
    $D$s. Only one of them will produce the run observed, so the probability is $1/66.$
  2. To have two runs, the two $D$s would either have to be at the beginning of the sequence,
    or at the end. There's only one way to do either, so this probability is $1/33.$
  3. We are given that there are two defectives, so the probability of having one run is zero.
    The probability of having two runs is $1/33.$ To have three runs, the two $D$s would have to be
    together, anywhere starting from the second position to the tenth. There are nine ways to do that,
    so we'd have $(2+9)/66=1/6$.

My Question for M.SE:

Note I'm making the assumption that two defectives are a given for all three parts of the question. I'm fairly sure that's correct. I know my answer for 3 is incorrect, but I don't know why; this leads me to suspect my answers for 1 and 2 are incorrect. Is there something amiss with my take on the "distinguishable versus undistinguishable" nature of the defectives? Or is there something else I'm missing?

Thank you for your time!

Best Answer

Questions 1 and 2 are correct.

You just forgot

DNNNNNNNNNND

That adds one more possibility on top of your 9+2 = 11 already counted, for a total of 12/66 or 2/11.

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