Retrieval check answer. By the same logic as the classic two-child problem: outcomes for two independent 50/50 sensors are {both bad, first-bad-only, second-bad-only, both fine}, each equally likely (25%). “At least one bad” rules out only “both fine,” leaving 3 equally likely outcomes, one of which is “both bad.” P(both bad | at least one bad) = 1/3 ≈ 0.333 — not 1/2. Knowing how the information arrived (a report that at least one is bad, not which one) changes the conditioning, same trap as Lesson 5.
Part 1 — Mode: 0. Highest single probability is 0.45, at “0 bugs remaining.”
Part 2 — Median: 1. Cumulative probability: P(0)=0.45, P(≤1)=0.75. The median is the smallest
value where cumulative probability reaches at least 0.5 — that’s crossed at “1 bug,” since 0.45 alone
falls short of 0.5 but 0.45+0.30=0.75 clears it.
Part 3 — Mean: 1.3.
0×0.45 + 1×0.30 + 2×0.10 + 3×0.10 + 10×0.05
= 0 + 0.30 + 0.20 + 0.30 + 0.50
= 1.30
Three genuinely different numbers from one posterior: mode 0, median 1, mean 1.3. This isn’t measurement noise or rounding — they’re answering three different questions, and a skewed distribution with a rare severe tail (the 5% chance of 10 bugs) is exactly where they’re guaranteed to disagree hardest. A symmetric, tail-free distribution would have all three coincide.
Part 4 — Report something closer to the mean, not the mode. The scenario states the penalty for under-preparing is worse than linear in bug count — a single 10-bug release is disproportionately costly, not just “10× as bad as a 1-bug release.” That’s a hint you’re closer to squared-error-shaped loss (or something even more tail-averse) than to flat absolute error, and the mean is exactly the number that accounts for how much probability sits in that expensive tail, weighted by how expensive it is. Reporting the mode (0 bugs) would tell the release manager “most likely, you’re completely clean” — true on its own terms, but it silently discards the 5% chance of the nightmare scenario entirely, precisely the scenario the manager most needs to plan around. The mode answers “what will probably happen,” not “what should I brace for” — and those are different questions whenever the downside is asymmetric. If the cost structure were flatter (a fixed, modest per-bug fix cost, no special penalty for clustering), the median would be the more defensible report; if you were being scored purely on nailing the exact headline number for a status report, the mode would be optimal. None of these numbers is “the truth” being approximated — each is the rational reply to a specific question about how you’re being scored for being wrong, and getting that mapping backwards (using the mode when your real losses are squared-error-shaped, say) means confidently reporting a minimize-the-wrong-thing summary.
The pattern:
| Estimate | Value | Minimizes | Good report when… |
|---|---|---|---|
| Mode | 0 | 0/1 loss (exact match only) | you’re scored on hitting the precise headline value |
| Median | 1 | absolute error | costs scale linearly with how wrong you are |
| Mean | 1.3 | squared error | large misses are disproportionately costly (this scenario) |
Where this goes: next lesson turns from single numbers to intervals — and a related, extremely common confusion: a Bayesian credible interval and a classical confidence interval can look identical on a page while claiming genuinely different things.