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Practice Solutions and Feedback

Use after a written attempt. If you read a solution first, mark that item as guided practice and use a changed case to assess independence.

Set A: models and conditioning​

A1. Protocol I retains seven equally likely sequences; only HHH qualifies, giving1/7. Protocol II gives selection likelihood1 to HHH,2/3 to each two-head sequence,1/3 to each one-head sequence, and0 to TTT. Total observation probability=.5, so posterior HHH=(1/8)/(.5)=1/4. Equal weights among retained sequences would be wrong for protocol II.

A2. Of200 expected defective requests,160 are flagged and40 missed. Of9,800 nondefective requests,392 are flagged and9,408 not flagged. Precision=160/552=20/69≈.289855; recall=.8; expected review volume552. Missing the9,800 denominator is a base-rate error.

A3. Union=.3+.4-.2=.5; conditional=.2/.4=.5. Independence would require intersection .3*.4=.12, so these events are dependent. Subtract overlap once because it was counted twice.

A4. Prior odds1/9; likelihood ratios4 and7; posterior odds28/9, so posterior28/37≈.756757. Equivalently numerator=.1*.8*.7=.056, alternative mass=.9*.2*.1=.018. Duplicate evidence does not provide a second independent likelihood ratio; use its conditional likelihood given the earlier record.

A5. Union at most1 implies intersection at least .8; divide by P(B)=.9 to get8/9. A table with both=.8,A-only=.1,B-only=.1,neither=0 achieves it. Asking for an arbitrarily small conditional probability with these marginals is impossible.

Set B: random variables​

B1. Fifteen pairs are equiprobable; defect counts0,1,2 have counts6,8,1. Mean=(8+2)/15=2/3; second moment=(8+4)/15=4/5; variance=4/5-4/9=16/45. Binomial variance is4/9, larger by4/45 because sampling without replacement induces negative dependence.

B2. Paths S,FS,FFS,FFFS,FFFF have probabilities .8,.16,.032,.0064,.0016. Expected attempts=1+.2+.04+.008=1.248; failures=.2+.04+.008+.0016=.2496; backoffs=.248; success=.9984. Expected waiting=12.4 ms. The final failed attempt is a failure but creates no further backoff.

B3. Integral of cx on[0,2] is2c, so c=.5. CDF on that interval is x²/4; probability below1 is1/4 and median is sqrt(2). Mean is integral x²/2 from0 to2, or4/3. Point probability is zero in this density model, not the height cx.

B4. Fifteen minutes=.25 hour, parameter lambda=2; zero-arrival probability=e^-2≈.135335. Exponential mean waiting=1/8 hour=7.5 minutes. A constant-rate Poisson process assumes independent increments and the relevant small-interval event behavior. Synchronized batch arrivals violate this simple model.

B5. X² takes0,1,4 with probabilities1/2,1/4,1/4. E[X]=1/4; E[X²]=5/4; Var(X)=5/4-1/16=19/16. Squaring E[X] gives1/16, not5/4.

Set C: inference​

C1. A particular server is empty if all four requests choose the other two, probability(2/3)^4. Linearity gives expected empty servers=3*(2/3)^4=16/27. The empty-server indicators need not be independent.

C2. Independent mean variance=.25/100=.0025. The copied average equals X exactly, so its variance=.25. Alternatively, the100 diagonal variance terms and9,900 ordered covariance terms each contribute .25 before dividing by10,000.

C3. Mean=1 and variance=1/n exactly. The CLT concerns the asymptotic distribution of sqrt(n)*(mean-1); it supplies no universally sufficient n=30. A highly skewed rare-event distribution can require far more observations.

C4. The intervals [0,0] and[1,1] never contain .5; coverage is0. Always reporting[0,1] covers every valid p but provides no precision. Judge width and coverage together under relevant models.

C5. Raw A=91,B=21.8 ms. At80/20 mix, A=28,B=31.6 ms. Counts and means do not provide within-group variances, dependence, or randomization information. They do not justify a confidence interval or causal effect. Collect unit-level measurements and an explicit assignment protocol.

C6. Simulation standard deviation=sqrt(.95*.05/5000)≈.003082. The .003 difference is about .97 standard deviations: compatible with ordinary simulation noise. It proves neither correctness nor a bug. Use exact enumeration and model checks alongside simulation.

For any corrected item, explain the original error in one sentence and solve a changed input without copying the derivation. The assessment page distinguishes guided practice from independent evidence.