Mock Midterm Exam #2

(a) Choose which 3 of the 10 positions hold a one:

\[\binom{10}{3} = \frac{10!}{3!\;7!} = 120\]

(b) We need to place 3 ones among 7 zeros such that no two ones are adjacent. Think of first placing the 7 zeros, which create 8 gaps (before the first zero, between consecutive zeros, and after the last zero):

\[\underset{\uparrow}{\_}\; 0 \;\underset{\uparrow}{\_}\; 0 \;\underset{\uparrow}{\_}\; 0 \;\underset{\uparrow}{\_}\; 0 \;\underset{\uparrow}{\_}\; 0 \;\underset{\uparrow}{\_}\; 0 \;\underset{\uparrow}{\_}\; 0 \;\underset{\uparrow}{\_}\]

Place the 3 ones in 3 distinct gaps (at most one per gap guarantees no adjacency):

\[\binom{8}{3} = 56\]

(a) With \(\lambda = 2.5\) for one day:

\[P(X = 0) = e^{-2.5} \approx 0.0821\]

(b) Over 2 days, \(\lambda_{2} = 2 \times 2.5 = 5\). We need \(P(X \geq 5) = 1 - P(X \leq 4)\):

\[P(X \leq 4) = e^{-5}\left(1 + 5 + \frac{25}{2} + \frac{125}{6} + \frac{625}{24}\right) = e^{-5}\left(1 + 5 + 12.5 + 20.833 + 26.042\right)\]

\[= e^{-5} \times 65.375 \approx 0.00674 \times 65.375 \approx 0.4405\]

\[P(X \geq 5) = 1 - 0.4405 \approx 0.560\]

There is roughly a 56% chance of seeing 5 or more failures in a 2-day window.

(a)

	Precision	Recall
Model A	\(\frac{6}{40} = 0.150\) (15%)	\(\frac{6}{8} = 0.750\) (75%)
Model B	\(\frac{5}{9} \approx 0.556\) (55.6%)	\(\frac{5}{8} = 0.625\) (62.5%)

(b) With a capacity of ≤10 reviews/day: Model B. It flags only 9 cases (within the daily budget) with much higher precision (55.6% vs 15%). Model A flags 40 cases — the team physically cannot review them all, meaning many alerts go uninvestigated anyway.

(c) With asymmetric costs (FN = €50,000, FP = €200): Model A. Compare expected costs:

• Model A: 2 missed frauds × €50,000 + 34 false alarms × €200 = €100,000 + €6,800 = €106,800
• Model B: 3 missed frauds × €50,000 + 4 false alarms × €200 = €150,000 + €800 = €150,800

When missing fraud is extremely costly, the extra recall of Model A (75% vs 62.5%) saves far more than the cost of its additional false alarms.

(a) Since \(X\) and \(Y\) are independent normals, \(D = X - Y\) is also normal:

\[E[D] = E[X] - E[Y] = 10 - 12 = -2\]

\[\text{Var}(D) = \text{Var}(X) + \text{Var}(Y) = 16 + 25 = 41\]

\[D \sim \mathcal{N}(-2,\; 41)\]

(b) \(P(X > Y) = P(D > 0)\). Since \(D \sim \mathcal{N}(-2,\; 41)\), we compute this directly from the CDF of that Gaussian:

\[P(D > 0) = 1 - \Phi_{\mathcal{N}(-2,\,41)}(0) \approx 0.378\]

There is roughly a 37.8% chance that Route X takes longer than Route Y, even though its mean is lower. The high variance on both routes leaves substantial overlap between the two distributions.

(a) Apply Bayes' theorem. First, compute \(P(\text{flagged})\) using the law of total probability:

\[P(\text{flagged}) = P(\text{flagged} \mid \text{phishing}) \cdot P(\text{phishing}) + P(\text{flagged} \mid \text{legit}) \cdot P(\text{legit})\]

\[= 0.96 \times 0.03 + 0.08 \times 0.97 = 0.0288 + 0.0776 = 0.1064\]

Now apply Bayes:

\[P(\text{phishing} \mid \text{flagged}) = \frac{P(\text{flagged} \mid \text{phishing}) \cdot P(\text{phishing})}{P(\text{flagged})} = \frac{0.0288}{0.1064} \approx 0.271\]

About 27.1% of flagged emails are actually phishing.

(b) The colleague is conflating sensitivity with precision. The filter has 96% sensitivity (it catches most phishing), but because phishing is rare (3% base rate), the vast majority of flagged emails are actually legitimate (false positives). Only about 27% of flagged emails are truly phishing — a user reviewing flagged emails would find roughly 3 out of 4 to be false alarms.

(a) Customer A's 90/100 performance translates to a prior (using a \(\text{Beta}(1,1)\) baseline):

\[\theta \sim \text{Beta}(1 + 90,\; 1 + 10) = \text{Beta}(91, 11)\]

Update with Customer B's pilot data (8 correct, 2 incorrect):

\[\theta \mid \text{data}_B \sim \text{Beta}(91 + 8,\; 11 + 2) = \text{Beta}(99, 13)\]

Posterior mean:

\[E[\theta] = \frac{99}{99 + 13} = \frac{99}{112} \approx 0.884\]

(b) Reporting "80%" ignores 100 observations from Customer A in the same domain. The pilot has only 10 data points — far too few to override what we already know. The posterior properly combines both sources of evidence, estimating accuracy near 88.4%, not 80%. Reporting raw pilot accuracy from a tiny sample is both noisy and wasteful of relevant prior information.

Choose 5 datasets for Team A from the 12, then 4 for Team B from the remaining 7, then the last 3 go to Team C:

\[\binom{12}{5} \times \binom{7}{4} \times \binom{3}{3} = 792 \times 35 \times 1 = 27{,}720\]

Equivalently, this is the multinomial coefficient \(\displaystyle\frac{12!}{5!\;4!\;3!} = 27{,}720\).

(a)

\[E[X] = 0(0.50) + 1(0.30) + 2(0.15) + 3(0.05) = 0 + 0.30 + 0.30 + 0.15 = 0.75\]

\[E[X^2] = 0^2(0.50) + 1^2(0.30) + 2^2(0.15) + 3^2(0.05) = 0 + 0.30 + 0.60 + 0.45 = 1.35\]

\[\text{Var}(X) = E[X^2] - (E[X])^2 = 1.35 - 0.75^2 = 1.35 - 0.5625 = 0.7875\]

(b) Let \(S = X_1 + X_2 + \cdots + X_{64}\). By the CLT:

\[E[S] = 64 \times 0.75 = 48 \qquad \text{Var}(S) = 64 \times 0.7875 = 50.4\]

\[\text{SD}(S) = \sqrt{50.4} \approx 7.10\]

So \(S \approx \mathcal{N}(48,\; 50.4)\). We want \(P(S > 55)\):

\[P(S > 55) = 1 - \Phi_{\mathcal{N}(48,\,50.4)}(55) \approx 0.162\]

About a 16.2% chance of finding more than 55 bugs across 64 reviews.