Original public practice · P-1 · October 2026

Probability: cumulative practice

Combine probability mechanisms, joint laws, limiting arguments and explicitly supplied process assumptions. This form contains 12 multipart questions and 120 practice points.

This is calculus-based undergraduate practice. Review sums, derivatives, integrals and conditional probability; statistics also uses vectors and matrices. Every lesson remains directly accessible. These original problems sample the published material; they are not an official university paper, a calibrated grade prediction or an examination of every course objective.

The process and information section supplies definitions and assumptions for extension tasks. Publishing these questions does not complete the remaining stochastic-process teaching sequence. For preparation, start with the linked probability lessons; attempt the three sections in separate sessions if helpful.

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For an optional rehearsal, allow 240 minutes across the form, or divide it into separate sessions with breaks and accommodations. This is an author estimate, not a measured learner timing. There is no enforced timer or pass threshold.

State the sampling model, parameter domain, units and theorem hypotheses. Distinguish exact finite-sample calculations, asymptotic approximations and empirical evidence. You may cite results taught in the linked lessons unless a part asks for their proof. Supply intermediate reasoning and the requested calculations; a theorem name or numerical match alone is not a proof. A calculator or local numerical code may help check arithmetic; no paid tool is required. Supplied theorem statements may be used only under their stated assumptions. Alternative valid methods can earn the same points.

Print this page for questions only. Solutions, review advice and navigation are excluded from print even when their disclosures are open. Keep a separate answer sheet.

1. Events and probability models

40 practice points.

Question 1: Sampling without replacement

10 points

A box contains four red and six blue distinguishable tokens. Select three tokens uniformly without replacement. Let R be the number of red tokens. A second experiment uses three independent draws with replacement from the original box.

(a) · 5 points

Derive the probability mass function of R, its support and P(R≥2). Explain why the masses sum to one without listing every sample.

(b) · 5 points

Use draw-position indicators to derive E(R) and Var(R). Compare the variance and P(R≥2) with replacement; explain the mechanism behind the difference.

Review question 1: worked argument and scoring criteria

(a)

Every unordered three-token subset has probability 1/C(10,3). Choosing r red and 3−r blue tokens gives P(R=r)=C(4,r)C(6,3−r)/120 for r=0,1,2,3. The numerators are 20,60,36,4. Thus P(R≥2)=40/120=1/3. The four disjoint count classes partition all three-token subsets, so their cardinalities sum to C(10,3).

2 points: Derives the count ratio with the correct support.
1 point: Obtains 1/3.
2 points: Justifies normalization by a disjoint exhaustive partition.

(b)

For indicators I₁,I₂,I₃, E(Iᵢ)=2/5 and Var(Iᵢ)=6/25. For distinct positions E(IᵢIⱼ)=(4/10)(3/9)=2/15, hence Cov(Iᵢ,Iⱼ)=−2/75. Therefore E(R)=6/5 and Var(R)=3(6/25)+6(−2/75)=14/25. With replacement R is Binomial(3,2/5), with variance 18/25 and tail 3(2/5)²(3/5)+(2/5)³=44/125. Removing a red token lowers the next red chance; independence is valid only in the replacement experiment.

2 points: Derives the negative covariance from a joint draw probability.
1 point: Computes the without-replacement mean and variance.
1 point: Computes both replacement comparisons.
1 point: Explains the dependence change.

Targeted lesson returns:

Question 2: Two signals and a selected sample

10 points

A population has disease probability 1/50. Given disease, two binary tests are independent with positive probability 9/10 each. Given no disease, they are independent with positive probability 1/10 each. These conditional independence assumptions are part of the model.

(a) · 5 points

Find the probability of two positive tests and the disease probability given two positives. Explain why multiplying the two unconditional positive probabilities is invalid.

(b) · 5 points

A clinic records a person if at least one test is positive. Find the disease probability in this selected sample. If only the two individual sensitivity/false-positive rates were supplied, could either posterior be identified? Give a reason or counterexample.

Review question 2: worked argument and scoring criteria

(a)

The two joint branches have probabilities (1/50)(81/100)=81/5000 and (49/50)(1/100)=49/5000. Thus P(++ )=130/5000=13/500 and P(D|++ )=81/130. A single positive has probability 29/250. Its square is not 13/500: the shared disease state creates unconditional dependence even though the tests are conditionally independent.

2 points: Builds and sums the correct joint branches.
2 points: Normalizes to 81/130.
1 point: Distinguishes conditional and marginal independence.

(b)

The selection probabilities are 1−(1/10)²=99/100 under disease and 1−(9/10)²=19/100 otherwise. Thus P(D|selected)=99/(99+49·19)=99/1030. Marginal rates alone do not specify test dependence. If the two tests are identical, both ++ and selection equal a single positive and both posteriors become (9/500)/(29/250)=9/58. This satisfies the same individual rates and changes both answers.

2 points: Computes both selection likelihoods.
1 point: Obtains 99/1030.
2 points: Supplies a valid alternative dependence model and its consequence.

Targeted lesson returns:

Question 3: An atom and a continuous response

10 points

With probability 1/4 a sensor returns X=0. Otherwise it returns an independent Uniform(0,2) reading. Define Y=X². All readings are in fixed normalized units.

(a) · 5 points

Give the complete CDF of X, P(X=0), P(X≤1), E(X) and Var(X). Explain why integrating the continuous density alone does not give total probability one.

(b) · 5 points

Derive the law of Y including its atom and density. Find P(Y>1) and E(Y). Does the divergent density near zero imply infinite probability or infinite mean?

Review question 3: worked argument and scoring criteria

(a)

F_X(x)=0 for x<0, 1/4+3x/8 for 0≤x<2, and 1 for x≥2. Thus P(X=0)=1/4 and P(X≤1)=5/8. E(X)=(3/4)·1=3/4 and E(X²)=(3/4)·(4/3)=1, so Var(X)=1−9/16=7/16. The continuous part has density 3/8 on (0,2) and mass 3/4; the missing 1/4 is an atom.

2 points: Gives the CDF including its jump and endpoints.
1 point: Computes the two requested probabilities.
1 point: Computes the mean and variance.
1 point: Accounts separately for the atom.

(b)

For 0≤y<4, F_Y(y)=P(X≤√y)=1/4+3√y/8. It is zero below zero and one at or above four. There is an atom 1/4 at zero and density 3/(16√y) on (0,4). Its integral is 3/4, P(Y>1)=3/8 and E(Y)=E(X²)=1. The density singularity is integrable; probability is its area, not its height. The bounded support also ensures a finite mean.

2 points: Derives support, atom and transformed density.
1 point: Checks the continuous mass.
1 point: Obtains the tail and mean.
1 point: Explains the integrable singularity.

Targeted lesson returns:

Question 4: Competing independent lifetimes

10 points

Independent lifetimes A and B are exponential with rates 2 and 3 per hour. Put T=min(A,B), and define the categorical label J as A when A<B and B when B<A, leaving it undefined on ties. Under the stated independent model ties have probability zero.

(a) · 5 points

Derive the survival function and expectation of T, and the probability that A fails first. Track the time units.

(b) · 5 points

Prove that T and J are independent by a joint event calculation. Explain what fails if instead A=B is a shared exponential lifetime.

Review question 4: worked argument and scoring criteria

(a)

For t≥0, P(T>t)=P(A>t)P(B>t)=e^(−5t), so T has rate 5 per hour and E(T)=∫₀∞e^(−5t)dt=1/5 hour. P(J=A)=∫₀∞2e^(−2t)e^(−3t)dt=2/5.

2 points: Uses independence to derive the survival law.
1 point: Derives the expectation with units.
2 points: Integrates the competing-event density to obtain 2/5.

(b)

For t≥0, P(T>t,J=A)=∫ₜ∞2e^(−5s)ds=(2/5)e^(−5t)=P(J=A)P(T>t). The B calculation gives (3/5)e^(−5t). These identities determine the conditional distribution of T for either label, establishing independence. With A=B, ties occur with probability one, survival cannot be multiplied, and J is undefined under the stated no-tie rule. Even if a tie rule is added, the minimum has the shared rate, not the sum of two rates.

3 points: Derives joint tails for both labels and uses them to prove independence.
2 points: Identifies the failed independence and no-tie assumptions in the changed model.

Targeted lesson returns:

2. Joint laws and limiting arguments

40 practice points.

Question 5: A sum and its fraction

10 points

X and Y are independent exponential variables of rate 1. Define S=X+Y and U=X/(X+Y). The event X+Y=0 has probability zero.

(a) · 5 points

Find the inverse transformation, its absolute Jacobian and the joint density of (S,U), including its exact support. Derive both marginals and decide independence.

(b) · 5 points

Now X has rate 2 and Y has rate 1, still independently. Derive f(S,U) and f_U. Decide whether S and U remain independent and compute P(X<Y).

Review question 5: worked argument and scoring criteria

(a)

The inverse is x=su,y=s(1−u), with s>0 and 0<u<1. The absolute determinant of [[u,s],[1−u,−s]] is s. Thus f(S,U)=se^(−s) on this support. Integrating gives f_S(s)=se^(−s), and f_U(u)=∫₀∞se^(−s)ds=1. Both integrate to one and the joint factors on a product support, so S and U are independent.

2 points: Gives inverse, support and Jacobian.
2 points: Derives and normalizes both marginals.
1 point: Justifies independence by factorization.

(b)

The same inverse gives f(S,U)=2s exp[−s(1+u)] for s>0,0<u<1. Integration in s gives f_U(u)=2/(1+u)², whose integral from 0 to 1 is 1. Given U=u, S has density (1+u)²s exp[−(1+u)s] and mean 2/(1+u), which varies with u; hence dependence. P(X<Y)=P(U<1/2)=∫₀^(1/2)2/(1+u)²du=2/3.

2 points: Obtains the changed joint and marginal densities.
2 points: Proves dependence using a conditional law or an equivalent argument.
1 point: Obtains 2/3.

Targeted lesson returns:

Derive a two-coordinate transformed law

Question 6: Prediction under a shared hidden rate

10 points

Let Θ be Uniform(0,1). Conditional on Θ=θ, observations X and Y are independent Bernoulli(θ) variables. Before observing X, predict Y with a constant; after observing X, predict Y using a function of X. Loss is squared error.

(a) · 5 points

Derive E(Y|X=1), E(Y|X=0) and Cov(X,Y). Is conditional independence enough to conclude marginal independence?

(b) · 5 points

Find the optimal constant and X-dependent predictions and their risks. Prove optimality among all real functions g(X) using conditional expectation.

Review question 6: worked argument and scoring criteria

(a)

P(X=1)=E(Θ)=1/2 and E(XY)=E(Θ²)=1/3. Thus E(Y|X=1)=(1/3)/(1/2)=2/3. Also P(Y=1,X=0)=E[Θ(1−Θ)]=1/6, giving E(Y|X=0)=1/3. Cov(X,Y)=1/3−1/4=1/12, so they are marginally dependent. Their shared random rate survives marginalization.

2 points: Computes the joint and conditioning denominators.
2 points: Obtains both conditional means.
1 point: Computes covariance and interprets the dependence.

(b)

The best constant is E(Y)=1/2 with risk Var(Y)=1/4. The conditional mean m(X) is 1/3 or 2/3. Both conditional Bernoulli variances equal 2/9, so its risk is 2/9, a reduction of 1/36. Write Y−g(X)=(Y−m(X))+(m(X)−g(X)). The cross term has expectation zero by conditioning on X, since E[Y−m(X)|X]=0. Therefore E[(Y−g(X))²]=2/9+E[(m(X)−g(X))²]≥2/9, with equality exactly when g=m almost surely.

2 points: Computes predictions, risks and reduction.
2 points: Derives the zero cross term and risk decomposition.
1 point: States the equality condition.

Targeted lesson returns:

Question 7: A random number of contributions

10 points

N is Poisson(λ), λ>0. Independently of N, Z₁,Z₂,… are iid with P(Z=1)=P(Z=2)=1/2. Let S=Σᵢ₌₁ᴺ Zᵢ, taking S=0 when N=0.

(a) · 5 points

Derive the moment-generating function of S by conditioning on N. State its domain and compute E(S) and Var(S), explaining why λVar(Z) alone is insufficient.

(b) · 5 points

Find P(S=2) by enumerating every possible N. Compare with a Poisson law of mean 3λ/2 and show that S is not that law for λ>0.

Review question 7: worked argument and scoring criteria

(a)

M_Z(t)=(e^t+e^(2t))/2 for every real t. Conditional on N=n, M_S(t|N=n)=M_Z(t)^n. Summing the Poisson series gives M_S(t)=exp{λ(M_Z(t)−1)}, finite for all real t. E(S)=λE(Z)=3λ/2. Total variance gives Var(S)=E[NVar(Z)]+Var[NE(Z)]=λ/4+9λ/4=5λ/2. Ignoring the second term omits the random number of summands.

2 points: Derives the MGF and its domain.
1 point: Computes the mean.
2 points: Computes both total-variance contributions.

(b)

S=2 occurs for N=1,Z₁=2, or N=2,Z₁=Z₂=1, with probabilities e^(−λ)λ/2 and e^(−λ)λ²/8. Thus P(S=2)=e^(−λ)(λ/2+λ²/8). A Poisson(3λ/2) would have mass e^(−3λ/2)(3λ/2)²/2 at 2. More decisively, S has P(S=0)=e^(−λ), whereas that Poisson law has e^(−3λ/2); these differ for every λ>0. Its unequal mean and variance provide another proof.

3 points: Enumerates exactly the two cases and their weights.
1 point: Gives the comparison mass.
1 point: Proves nonidentity for every positive λ.

Targeted lesson returns:

Question 8: A guarantee, an approximation and a counterexample

10 points

X₁,…,Xₙ are iid Bernoulli(p), where 0≤p≤1 is unknown, and X̄ is their average. You may use the iid finite-variance CLT: for fixed 0<p<1, √n(X̄−p)/√[p(1−p)] converges in distribution to a standard normal. A standard normal has probability approximately 0.95 between −1.96 and 1.96.

(a) · 5 points

Use Chebyshev to choose n guaranteeing P(|X̄−p|≥0.05)≤0.05 for every p. For fixed p=1/2, give the corresponding normal-based sample-size estimate and explain why it is not the same guarantee.

(b) · 5 points

Replace the iid observations by Xᵢ=Z for all i, with Z Bernoulli(1/2). Compute the same tail probability for any n. Separately explain what happens to the CLT normalization at p=0 or p=1 in the original model.

Review question 8: worked argument and scoring criteria

(a)

Var(X̄)=p(1−p)/n≤1/(4n). Chebyshev gives a tail bound 1/[4n(0.05)²]=100/n, so n≥2000 suffices uniformly. The normal heuristic at p=1/2 asks 0.05√n/0.5≥1.96, giving n≥384.16, hence 385 after rounding. The supplied CLT is an asymptotic statement for fixed p, with no finite-n error bound, so 385 is not certified by that theorem to meet the exact tail requirement.

2 points: Derives the uniform variance/tail bound.
1 point: Obtains 2000.
1 point: Obtains the approximate 385.
1 point: Distinguishes finite-sample guarantee from asymptotics.

(b)

Now X̄=Z regardless of n, so |X̄−1/2|=1/2 with probability one and the tail probability is 1. The individual marginals still have variance 1/4, but covariance terms prevent Var(X̄) from shrinking. At p=0 or 1 in the iid model the average is exactly p and the deviation probability is zero; the displayed CLT ratio is undefined because the denominator is zero, so the degenerate case is handled directly.

2 points: Computes the perfectly dependent tail and variance behavior.
1 point: Identifies the lost independence assumption.
2 points: Handles the two degenerate endpoints without dividing by zero.

Targeted lesson returns:

3. Processes and information

40 practice points.

Question 9: Equal count marginals do not define a process

10 points

For this question, a rate-2 Poisson process N(t), t≥0 in hours, has N(0)=0, independent counts on disjoint time intervals and increment N(t+h)−N(t) distributed Poisson(2h). Its arrival time T₁ satisfies {T₁>t}={N(t)=0}. These are supplied model assumptions, not consequences of a single Poisson marginal.

(a) · 5 points

Find P(N(1)=1,N(3)=4). Derive P(N(1)=1|N(3)=4), explicitly identifying the independent quantities.

(b) · 5 points

Derive the law of T₁. Then let C₁=C₂=Z, where Z is Poisson(2). Each one-hour count has the same marginal as the process. Compare P(C₁=C₂=0) with independent consecutive process counts, and explain why the marginals do not identify a Poisson process.

Review question 9: worked argument and scoring criteria

(a)

N(1) and N(3)−N(1) are independent Poisson(2) and Poisson(4). The joint probability is (2e^(−2))(4³e^(−4)/3!)=(64/3)e^(−6). Divide by P(N(3)=4)=6⁴e^(−6)/4!=54e^(−6) to get 32/81. Equivalently, the conditional count is Binomial(4,1/3), giving 4(1/3)(2/3)³. N(1) and N(3) themselves are not independent.

2 points: Uses the correct disjoint increments.
1 point: Obtains the joint probability.
1 point: Obtains 32/81.
1 point: Rejects independence of the nested counts.

(b)

P(T₁>t)=P(N(t)=0)=e^(−2t), t≥0, so T₁ is exponential of rate 2 and mean 1/2 hour. For the shared-count model P(C₁=C₂=0)=P(Z=0)=e^(−2); independent process counts give e^(−4). Thus identical one-hour marginal distributions can have different joint laws. The shared-count construction is a pair of interval counts, not a full continuous-time construction; it already disproves identification from those marginals alone.

2 points: Derives survival, rate and mean.
2 points: Computes both zero-count probabilities.
1 point: States the precise limit of marginal information.

Targeted lesson returns:

Question 10: First-step equations and eventual absorption

10 points

A time-homogeneous Markov chain has states A,B,C,D. From A it moves to B with probability 1/2 and C with probability 1/2. From B it moves to A with probability 1/4, C with probability 1/4 and D with probability 1/2. C and D are absorbing. Conditional on the current state, future transition probabilities do not depend on earlier states. Let T be the number of transitions until first reaching C or D.

Transition probabilities; rows are current states
From / to	A	B	C	D
A	0	1/2	1/2	0
B	1/4	0	1/4	1/2
C	0	0	1	0
D	0	0	0	1

(a) · 5 points

Prove absorption occurs almost surely and that E(T) is finite starting from A or B. Then derive and solve first-step equations for the probability of absorption at D.

(b) · 5 points

Compute E_A(T) and E_B(T). Now change only the B row to B→A with probability one. Recompute the D-absorption probability and expected time from A; explain why retaining the old equations is invalid.

Review question 10: worked argument and scoring criteria

(a)

From either transient state the probability of remaining transient for one more transition is at most 1/2. Conditional iteration gives P(T>n)≤(1/2)^n. Hence P(T=∞)=0 and E(T)=Σₙ≥0P(T>n)≤2. Write d_C=0,d_D=1, d_A=d_B/2 and d_B=d_A/4+1/2. Solving gives d_B=4/7 and d_A=2/7. The absorption argument ensures that these events exhaust the eventual outcomes.

2 points: Proves the geometric tail bound and finite expectation.
2 points: Sets up equations with correct absorbing boundaries.
1 point: Solves both probabilities.

(b)

The original equations are t_A=1+t_B/2 and t_B=1+t_A/4, giving t_A=12/7 and t_B=10/7. After the change, D is unreachable from A or B, so its absorption probability is zero. Each two-transition block starting at A survives with probability 1/2, so absorption at C remains almost sure with finite expectation. The new equations t_A=1+t_B/2,t_B=1+t_A give t_A=3,t_B=4. Transition changes alter the future-state mixture while each transition still costs one step; the original B equation no longer describes the model.

2 points: Obtains both original mean times.
1 point: Identifies the unreachable D outcome.
2 points: Justifies finiteness and solves the changed mean-time equations.

Targeted lesson returns:

Question 11: Expected code length and model mismatch

10 points

A source produces independent symbols a,b,c with probabilities p=(1/2,1/4,1/4). For a positive probability vector q, define ideal length ℓ_q(i)=−log₂qᵢ and entropy H(p)=−Σpᵢlog₂pᵢ. Ideal lengths need not be integer bit lengths. You may use ln x≤x−1 for x>0, with equality only at x=1.

(a) · 5 points

Compute H(p). Compare the expected ideal length under q=(1/3,1/3,1/3) with H(p). Give actual binary prefix codewords (no codeword is the beginning of another) attaining H(p) for this p, and explain why this does not implement a general file compressor.

(b) · 5 points

For any strictly positive finite probability vectors p and q, prove E_p(ℓ_q)≥H(p) and characterize equality using the supplied logarithm inequality. What if q assigns zero to a symbol of positive p?

Review question 11: worked argument and scoring criteria

(a)

H(p)=(1/2)·1+2(1/4)·2=3/2 bits per symbol. Uniform q gives log₂3 bits per symbol, with excess log₂3−3/2≈0.08496. Codewords a→0,b→10,c→11 are prefix-free and have expected length 3/2. This special dyadic model admits integer ideal lengths; arbitrary distributions, metadata, finite files and model fitting introduce issues not solved by this example.

2 points: Computes entropy and uniform expected length.
1 point: Computes their difference.
1 point: Provides and verifies a prefix code.
1 point: States the implementation boundary.

(b)

The excess is Σpᵢlog₂(pᵢ/qᵢ). Apply ln(qᵢ/pᵢ)≤qᵢ/pᵢ−1, multiply by pᵢ and sum: Σpᵢln(qᵢ/pᵢ)≤Σqᵢ−Σpᵢ=0. Negating and dividing by ln2 proves nonnegativity. Since every pᵢ>0, equality holds exactly when qᵢ/pᵢ=1 for every i, namely p=q. If some qᵢ=0<pᵢ, that symbol has infinite ideal length and the expected length is infinite, so the finite logarithmic calculation cannot be used unchanged.

3 points: Derives and sums the logarithm inequality with correct sign and base.
1 point: Proves the equality condition.
1 point: Handles a zero model probability explicitly.

Targeted lesson returns:

Question 12: A bounded stopping rule

10 points

ξ₁,ξ₂,… are independent signs, each ±1 with probability 1/2. Set S₀=0 and Sₙ=ξ₁+⋯+ξₙ. Let τ be the first n with Sₙ=2, or infinity if no such n exists. Set T=min(τ,3). The available history at step n is ξ₁,…,ξₙ. For a finite bound m, you may use the pathwise identity S_T=Σⱼ₌₁ᵐ ξⱼ·1{T≥j} when T≤m. Do not assume an optional-stopping theorem for unbounded times.

(a) · 5 points

Enumerate the distribution of S_T and compute its expectation. Explain why the event T≥j is determined before ξⱼ is observed.

(b) · 5 points

Prove E(S_T)=0 for any stopping time bounded by m using the supplied identity and independence. If the cap is removed, explain why this proof does not establish E(S_τ)=0, even if eventual hitting is asserted separately.

Review question 12: worked argument and scoring criteria

(a)

Two initial positive signs hit 2 at time 2 with probability 1/4, giving S_T=2. The other six length-three sign paths have probabilities 1/8 each: +−+ and −++ end at 1; +−−,−+−,−−+ end at −1; −−− ends at −3. Thus probabilities at 2,1,−1,−3 are 1/4,1/4,3/8,1/8, with expectation 1/2+1/4−3/8−3/8=0. T≥j means no stop before j; that is determined by the signs up to j−1, including the fixed cap.

3 points: Enumerates all stopped outcomes with their probabilities.
1 point: Computes zero expectation.
1 point: Identifies the pre-increment information for survival.

(b)

For each j≤m, the indicator 1{T≥j} is determined by the past and independent of ξⱼ, so E[ξⱼ1{T≥j}]=E(ξⱼ)P(T≥j)=0. Linearity for the finite sum gives E(S_T)=0. Removing the cap creates an infinite sum or a limit of stopped sums; exchanging expectation with that limit requires an integrability or uniform-integrability argument, which has not been supplied. If τ is finite almost surely, then S_τ=2 almost surely, so its expectation is 2, not 0. This conditional observation diagnoses the invalid limit step; it is not a proof of almost-sure hitting.

3 points: Proves zero expectation using past measurability, independence and a finite sum.
1 point: Identifies the unjustified expectation-limit interchange.
1 point: Correctly interprets the hypothetical eventual-hitting conclusion.

Targeted lesson returns:

Turn the review into a new argument

Award points for reasoning actually present. Allow valid later reasoning based on a carried-forward value unless its criterion requires the correct value; do not award correct-result points for an incorrect result. Record missing hypotheses, invalid conditioning, unsupported calibration and arithmetic errors separately.

The total is out of 120, with no university-grade conversion. Preserve your first attempt, consult the targeted lesson, explain the repair and attempt a fresh case. A teacher or peer may review the complete argument; if nobody did, label it self-review. Public worked solutions cannot certify independent proficiency.

Return to Probability: cumulative practice for focused practice. The sustained probability investigation supplies a separate report and changed-case task. Independent assessment remains a separate requirement.

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