Probability & statistics

Conditioning changes the denominator. Derive the action threshold by comparing expected losses, not by choosing a conventional probability cutoff.

The first population has TP 95,FN 5,FP 495,TN 9405. P(defect|flag)=95/590≈16.10%; P(defect|no flag)=5/9410≈0.0531%. Sensitivity is 95/100=95%, specificity 9405/9900=95%. With 20% prevalence in 10,000 items, TP 1900,FN 100,FP 400,TN 7600, so positive predictive value is 1900/2300≈82.61%. Sensitivity and specificity were held fixed; the prior probability changed.

Replacing has expected loss 4(1−p), while accepting has 60p. Replace when 4(1−p)<60p, giving p>1/16=6.25%, with indifference at equality. Under the first population's assumptions a positive flag exceeds this threshold, while a negative flag does not. Different costs can change that policy even with identical detector accuracy. A binary output cannot rank risks within its positive and negative groups. Transferring the detector to another population also requires checking, rather than assuming, that its conditional error rates remain stable.

Check your reasoning

• Make all four cells sum to the stated population.
• Use the flagged denominator for precision.
• State both the prevalence assumption and the cost model.

Marginal failure includes common-power intervals. Use the law of total probability before multiplying conditional probabilities.

With independence, joint unavailability is 0.02²=0.0004. In the common-cause model,0.01+0.99q=0.02, so q=1/99≈0.010101. Joint unavailability is 0.01+0.99q²=1/99≈0.010101, about 25.25 times the independent result.

In both models E[X+Y]=E[X]+E[Y]=0.04; linearity of expectation requires no independence. Each marginal variance is 0.02×0.98=0.0196. Independence gives total variance 0.0392. With shared power, Cov(X,Y)=E[XY]−E[X]E[Y]=1/99−0.0004≈0.009701, so variance is 0.0392+2×0.009701≈0.058602.

The components' identical individual reliability does not establish the joint behavior needed for service availability. Independence is a structural assumption about failure causes. Adding replicas inside one failure domain may add much less protection than a multiplication of marginal probabilities suggests.

Check your reasoning

• Solve q from the marginal requirement.
• Separate unconditional and conditional independence.
• Explain why expectation is unchanged while variance and joint failure change.

The standard error uses the square root of sample size. A bound for missing observations is a different calculation from a confidence interval.

The estimated standard error is 4/√100=0.4 ms. The approximate interval is 12±1.96×0.4, or[11.216,12.784] ms. It assumes a sampling scheme and observations for which this approximation is appropriate, including sufficient independence and a suitable distribution of the sample mean. Across repeated samples, the method aims for 95% coverage; after observing this interval the fixed population mean is not assigned a frequentist 95% probability by this calculation alone.

The observed total is 1200 ms. Missing latencies contribute between 0 and 800 ms, so the combined mean is bounded by[1200/120,2000/120]=[10,16.6667] ms. These are identification bounds under the supplied range, not 95% confidence limits. In a real timeout system the timeout threshold could impose a stronger lower bound, but it was not given here.

A larger sample from the same selected population may estimate the mean conditional on completion more precisely. It does not make that conditional mean equal the mean of all requests. Sampling uncertainty, measurement error, and selection bias require separate arguments.

Check your reasoning

• Label the interval as approximate and state its sampling assumptions.
• Compute bounds from totals, including the missing observations.
• Do not describe a confidence interval as a cure for selection bias.

Changing subgroup weights changes the pooled rate. For the randomized trial use SE=sqrt(p1(1−p1)/n1+p0(1−p0)/n0).

The new interface rates are 90% and 20%; the old rates are 95% and 50%. Pooled rates are 94/120≈78.33% versus 69/120=57.5%, apparently favoring new despite worse performance in both subgroups. Equal-weight standardized rates are 0.5×0.90+0.5×0.20=55% and 0.5×0.95+0.5×0.50=72.5%. Assignment and user experience are entangled, so the pooled association cannot isolate the interface effect. Even adjustment requires assumptions about other differences.

Randomize within experience strata, record allocation before outcomes, use the same task and observation window, and predeclare the outcome and stopping rule. For the separate randomized data the difference is 0.52−0.50=0.02. SE=√(0.2496/1000+0.25/1000)≈0.02235, giving approximately[−0.02381,0.06381]. The data are compatible with both a small decrease and a modest increase at this precision. The interval is not a probability that the effect lies inside, and failure to exclude zero does not prove equivalence.

Check your reasoning

• Calculate rates within groups before pooling.
• Use the specified common weights for standardization.
• Separate a causal identification problem from statistical imprecision.

Let an unobserved common cause determine both variables. Predicting one variable from another does not identify the effect of intervening on it.

The fitted line is a=0,b=10, with all residuals zero. One compatible model has region population z=1,2,3 million, advertising budget x=z, and revenue y=10z. In that model changing x while holding z fixed does not change y. Another model could include an advertising effect. The observations alone cannot distinguish those causal structures.

An experiment would randomly assign an advertising change across comparable geographic units, define the revenue window, measure spillovers and baseline differences, and estimate effects at the assigned intervention level. Too few independent regions can make uncertainty large even with many customers per region.

A held-out dataset generated by the same population-to-budget policy can preserve the relationship y≈10x and yield excellent predictive accuracy. Such a split tests prediction under that distribution, not the response to a changed budgeting policy. Extrapolating beyond observed budgets introduces an additional assumption. Report the fitted association and the intervention question separately.

Check your reasoning

• Calculate the line and residuals explicitly.
• Give a complete alternative mechanism consistent with the data.
• Identify the unit of randomization and distinguish prediction from intervention.

Multiply g by the conjugate. The square-root operation and intermediate x² still require range analysis.

Multiplying numerator and denominator by sqrt(x²+1)+x gives g(x)=1/(sqrt(x²+1)+x). At x=10^8, binary64 may round x²+1 to x², making the direct subtraction zero. The rearranged expression gives about 5×10^−9, matching the leading approximation 1/(2x). Algebraic equality over real numbers does not imply identical floating-point evaluation.

For much larger x, forming x² can overflow even when the final answer is representable. A range-safe hypotenuse computation, such as a suitably implemented hypot(x,1), addresses that intermediate risk; the sum in the denominator also needs consideration near the largest representable values.

Under usual binary64 rounding, (10^16+1)−10^16 is 0 because the addition loses 1. The second expression is also 0 in this particular ordering, so it is not a useful associativity counterexample. Compare instead (10^16−10^16)+1=1 with 10^16+(−10^16+1)=0. A careful numerical critique checks the proposed counterexample rather than repeating a slogan. Instability describes unnecessary error introduced by an algorithm; conditioning describes sensitivity of the mathematical answer to small input perturbations.

Check your reasoning

• Derive the conjugate form and test both implementations.
• Separate cancellation from overflow.
• Verify the actual floating-point example before claiming non-associativity.