Exercise 3: A Defect Batch, and a Broken Independence Assumption — Possible Solution
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GIVEN
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n = 20, k = 1, p = 0.05

STEP 1: P(EXACTLY 1 DEFECTIVE UNIT)
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C(20, 1) = 20
p^1 = 0.05
(1-p)^(20-1) = 0.95^19 ~= 0.377

P(X=1) = 20 x 0.05 x 0.377
        ~= 0.3774 (about 37.74%)

STEP 2: E[X]
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E[X] = n x p = 20 x 0.05 = 1.0

The expected number of defective units in the batch is exactly 1.

STEP 3: WHICH CONDITION WOULD BE VIOLATED
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If a single faulty machine setting affects every unit in the batch
identically, the "independent trials" condition from this chapter's
own four requirements is the one that breaks. A shared root cause
means the units are no longer failing independently of one another -
if the machine setting is bad, many or most units are likely to be
defective together, and if it's fine, few or none are. This is
exactly the same kind of dependency Chapter 3 warned about generally,
now shown breaking a specific distribution's own required assumptions.

THE PRACTICAL CONSEQUENCE
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Per this chapter's own warning box, treating correlated defects as if
they were independent will systematically UNDERESTIMATE how often
extreme outcomes occur - the binomial model would predict that a
batch with, say, 15 defective units is astronomically unlikely (the
same way a zero-defect batch of 500 units was shown to be
astronomically unlikely in this chapter's own worked example), when
in reality a single bad machine setting could make "almost the whole
batch is defective" a routine, entirely plausible outcome rather than
a near-impossible one.

WHY THIS WORKS AS AN ANSWER
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The probability and mean are computed directly using this chapter's
own binomial formula and shortcut, and the independence-violation
question is answered by naming the specific one of this chapter's own
four required conditions that a shared root cause breaks, then
explaining the real consequence using this chapter's own extreme-
outcome-underestimation warning rather than a generic "the model would
be wrong" statement.