Exercise 1: A Different Rollout's Clean-Session Rate and Independence Check — Possible Solution
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GIVEN
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P(A) = 0.04, P(B) = 0.025, P(A and B) = 0.001

STEP 1: P(CLEAN SESSION)
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Per Step 1's own union rule:

  P(A or B) = P(A) + P(B) - P(A and B)
            = 0.04 + 0.025 - 0.001
            = 0.064

Per Step 1's own complement rule:

  P(clean session) = 1 - P(A or B) = 1 - 0.064 = 0.936 (93.6%)

STEP 2: THE INDEPENDENCE CHECK
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Per Step 2's own multiplication-rule test:

  P(A) x P(B) = 0.04 x 0.025 = 0.001

Comparing to the actual P(A and B) = 0.001 - the two values match
exactly. Unlike the main worked example (where the two problem types
were clearly dependent, clustering together far more than chance would
predict), A and B here ARE independent: knowing one problem occurred
tells you nothing extra about whether the other one did too.

WHY THIS WORKS AS AN ANSWER
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Both calculations reuse this chapter's own Step 1 and Step 2 formulas
exactly, and the independence conclusion is drawn by directly comparing
P(A)xP(B) against the actual joint probability rather than assuming
dependence just because the main worked example happened to find it -
this dataset genuinely produces the opposite result.