Mock Midterm Exam — Solutions

We know \(P(\text{AID} \mid \text{KM}) = 0.28\) and \(P(\text{KM}) = 0.35\).

By the definition of conditional probability:

\[P(\text{KM} \cap \text{AID}) = P(\text{AID} \mid \text{KM}) \cdot P(\text{KM}) = 0.28 \times 0.35 = 0.098\]

Using Bayes' theorem:

\[P(\text{KM} \mid \text{AID}) = \frac{P(\text{AID} \mid \text{KM}) \cdot P(\text{KM})}{P(\text{AID})} = \frac{0.28 \times 0.35}{0.40} = \frac{0.098}{0.40} = 0.245\]

Each of the 11 positions can independently be one of 4 nucleotides. By the multiplication rule:

\[4^{11} = 4{,}194{,}304\]

First, choose 3 students out of 99 for the first group, then 3 out of 96 for the second, and so on:

\[\binom{99}{3}\binom{96}{3}\binom{93}{3}\cdots\binom{3}{3} = \frac{99!}{(3!)^{33}}\]

However, this treats the 33 groups as distinguishable (first group chosen, second group chosen, etc.). Since the order of the groups does not matter, we divide by \(33!\):

\[\frac{99!}{(3!)^{33} \cdot 33!}\]

First, satisfy the minimums: give $1M to C1, $2M to C2, $3M to C3, $4M to C4. That uses \(1+2+3+4 = 10\) million, leaving \(20 - 10 = 10\) million to distribute freely among 4 companies.

This is a stars-and-bars problem: distribute 10 identical units among 4 bins (each bin \(\geq 0\)):

\[\binom{10 + 4 - 1}{4 - 1} = \binom{13}{3} = 286\]

We need to count allocations where at least 3 companies receive their minimum (or more). We split into two cases:

Case 1: All 4 companies receive investment (exactly the answer from part (a)):

\[\binom{13}{3} = 286\]

Case 2: Exactly 3 companies receive investment (one company gets $0). We skip each company in turn and distribute $20M among the remaining 3, respecting their minimums:

• Skip C1 (min $1): remaining minimums \(2+3+4=9\), leftover \(20-9=11\) among 3 bins → \(\binom{11+2}{2} = \binom{13}{2} = 78\)
• Skip C2 (min $2): remaining minimums \(1+3+4=8\), leftover \(20-8=12\) among 3 bins → \(\binom{12+2}{2} = \binom{14}{2} = 91\)
• Skip C3 (min $3): remaining minimums \(1+2+4=7\), leftover \(20-7=13\) among 3 bins → \(\binom{13+2}{2} = \binom{15}{2} = 105\)
• Skip C4 (min $4): remaining minimums \(1+2+3=6\), leftover \(20-6=14\) among 3 bins → \(\binom{14+2}{2} = \binom{16}{2} = 120\)

Case 2 total: \(78 + 91 + 105 + 120 = 394\)

Final answer:

\[286 + 394 = 680\]

First, estimate the rate parameter. Over 400 hours we observed 2 events:

\[\lambda = \frac{2}{400} = 0.005 \text{ events/hour}\]

For a 3-hour window, the expected number of events is:

\[\mu = \lambda \cdot t = 0.005 \times 3 = 0.015\]

In a Poisson process, \(P(X = k) = \frac{\mu^k e^{-\mu}}{k!}\). We want at least one occurrence:

\[P(X \geq 1) = 1 - P(X = 0) = 1 - e^{-0.015} \approx 1 - 0.9851 = 0.0149\]

About a 1.5% chance the bug strikes during a 3-hour task.

Since Customer B is in the same domain as Customer A, we can use the Customer A results as an informative prior. The 85/100 performance on Customer A translates to a prior:

\[\theta \sim \text{Beta}(85, 15)\]

Now we update with Customer B's data (6 successes, 4 failures):

\[\theta \mid \text{data}_B \sim \text{Beta}(85 + 6,\; 15 + 4) = \text{Beta}(91, 19)\]

The posterior mean is:

\[\mathbb{E}[\theta] = \frac{91}{91 + 19} = \frac{91}{110} \approx 0.827\]

So rather than reporting the naive 6/10 = 60% from Customer B's small test alone, we report a posterior estimate of about 82.7%. The strong prior from the same domain pulls the estimate up, reflecting that 10 data points are too few to override what we already know about the model's performance in online game sales.

We should also report uncertainty. A 95% credible interval is the interval \([L, U]\) such that 95% of the posterior probability lies between \(L\) and \(U\). For a \(\text{Beta}(\alpha, \beta)\) posterior, this is computed as the interval between the 0.025 and 0.975 quantiles of that distribution — meaning we believe with 95% posterior probability that the true parameter falls within this range.

The 95% credible interval for \(\text{Beta}(91, 19)\) is approximately [0.75, 0.89].

If Customer B is in a different domain (music vs games), we should not use the Customer A results as a strong prior, since the model's accuracy in game sales may not transfer to music sales.

Instead, we should use a weakly informative or uninformative prior, for example \(\text{Beta}(1, 1)\) (uniform):

\[\theta \sim \text{Beta}(1, 1)\]

Updating with Customer B's data:

\[\theta \mid \text{data}_B \sim \text{Beta}(1 + 6,\; 1 + 4) = \text{Beta}(7, 5)\]

The posterior mean is:

\[\mathbb{E}[\theta] = \frac{7}{12} \approx 0.583\]

Now the estimate (≈ 58.3%) is much lower and much more uncertain. The 95% credible interval for \(\text{Beta}(7, 5)\) is approximately [0.31, 0.83] — very wide, reflecting that we have only 10 observations and no relevant prior information.

Key insight: The choice of prior matters. Using domain knowledge appropriately (same domain → informative prior, different domain → weak prior) is a critical modeling decision that directly affects the accuracy estimate and its uncertainty.

Both teams have roughly equally reliable test estimates — Team A's are slightly more reliable (200 test examples vs 180).

The reliability of an accuracy estimate depends on the absolute size of the test set, not the total dataset size. The standard error of an estimated proportion \(\hat{p}\) is:

\[\text{SE} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n_{\text{test}}}}\]

Only \(n_{\text{test}}\) appears in this formula — nothing about the training set or the total dataset. Team A tests on 200 examples and Team B on 180, so their standard errors are very similar.

With a \(\text{Beta}(1,1)\) prior (uniform) and Binomial data, the posterior is:

\[\theta \mid \text{data} \sim \text{Beta}(1 + k,\; 1 + n - k)\]

Team A (120 correct out of 200):

\[\theta_A \sim \text{Beta}(121, 81)\]

Team B (175 correct out of 180):

\[\theta_B \sim \text{Beta}(176, 6)\]

The variance of a \(\text{Beta}(\alpha, \beta)\) distribution is:

\[\text{Var} = \frac{\alpha \beta}{(\alpha+\beta)^2(\alpha+\beta+1)}\]

Team A:

\[\text{Var}_A = \frac{121 \times 81}{202^2 \times 203} = \frac{9801}{8{,}283{,}212} \approx 0.001183\]

Team B:

\[\text{Var}_B = \frac{176 \times 6}{182^2 \times 183} = \frac{1056}{6{,}061{,}692} \approx 0.000174\]

Team A's posterior has the wider 95% credible interval.

The Beta distribution's variance depends on how far the posterior mean is from 0.5. Team A's posterior mean is \(\frac{121}{202} \approx 0.60\), close to the high-uncertainty zone. Team B's posterior mean is \(\frac{176}{182} \approx 0.967\), far toward 1, where the distribution is tightly concentrated. So even though both teams have similar test set sizes, Team A's estimate of accuracy is considerably less precise.

(a) We compute \(E[X]\) directly:

\[E[X] = 0(0.10) + 1(0.20) + 5(0.35) + 10(0.30) + 20(0.05) = 0 + 0.20 + 1.75 + 3.00 + 1.00 = \$5.95\]

(b) First compute \(E[X^2]\):

\[E[X^2] = 0^2(0.10) + 1^2(0.20) + 5^2(0.35) + 10^2(0.30) + 20^2(0.05) = 0 + 0.20 + 8.75 + 30.00 + 20.00 = 58.95\]

\[\text{Var}(X) = E[X^2] - (E[X])^2 = 58.95 - (5.95)^2 = 58.95 - 35.4025 = 23.5475\]

(c) Let \(S = X_1 + X_2 + \cdots + X_{50}\). Since the donations are IID with \(n = 50\), by the CLT:

\[S \approx \mathcal{N}(\mu_S,\, \sigma^2_S) \quad \text{where} \quad \mu_S = 50 \times 5.95 = 297.50,\quad \sigma^2_S = 50 \times 23.5475 = 1177.375\]

So \(S \approx \mathcal{N}(297.50,\; 1177.375)\).

(d) Since \(S \sim \mathcal{N}(297.50,\; 1177.375)\), we can directly compute:

\[P(S \geq 350) = 1 - P(S < 350) \approx 0.063\]

Assumption: \(X\) and \(Y\) are independent.

A linear combination of independent normal random variables is also normal. We compute the mean and variance:

\[E[2X + Y] = 2E[X] + E[Y] = 2(1) + 1 = 3\]

\[\text{Var}(2X + Y) = 4\cdot\text{Var}(X) + \text{Var}(Y) = 4(2) + 2 = 10\]

Therefore: \(2X + Y \sim \mathcal{N}(3,\; 10)\).

"At least 80 rolls are necessary" means that after 79 rolls the sum has not yet exceeded 300, i.e. \(S_{79} \leq 300\).

For a fair die, each roll \(X_i\) has:

\[E[X] = \frac{7}{2} = 3.5 \qquad \text{Var}(X) = \frac{35}{12} \approx 2.9167\]

Let \(S_{79} = X_1 + \cdots + X_{79}\). By the CLT:

\[E[S_{79}] = 79 \times 3.5 = 276.5 \qquad \text{Var}(S_{79}) = 79 \times \frac{35}{12} \approx 230.417 \qquad \sigma_{S_{79}} \approx 15.18\]

Since \(S_{79} \sim \mathcal{N}(276.5,\; 230.417)\), we can directly compute:

\[P(S_{79} \leq 300) \approx 0.9392\]

There is approximately a 93.9% chance that at least 80 rolls are needed.

(a) Let \(S_A = A_1 + \cdots + A_{20}\). By the CLT:

\[E[S_A] = 20 \times 50 = 1000 \qquad \text{Var}(S_A) = 20 \times 100 = 2000\]

So \(S_A \sim \mathcal{N}(1000,\; 2000)\), and we directly compute \(P(S_A < 950) \approx 0.1318\).

(b) Let \(S_B = B_1 + \cdots + B_{20}\). By the CLT:

\[E[S_B] = 20 \times 52 = 1040 \qquad \text{Var}(S_B) = 20 \times 200 = 4000\]

So \(S_B \sim \mathcal{N}(1040,\; 4000)\), and we directly compute \(P(S_B < 950) \approx 0.0774\).

(c) Consider \(D = S_A - S_B\). Assuming independence between programs:

\[E[D] = 1000 - 1040 = -40 \qquad \text{Var}(D) = 2000 + 4000 = 6000\]

So \(D \sim \mathcal{N}(-40,\; 6000)\), and we directly compute \(P(D < 0) \approx 0.6972\).

There is approximately a 69.7% chance that Program A finishes before Program B.