Based on my handwritten notes below and my tutoring session
transcript (session_transcript.Rmd, included in this
submission).
Estimated time: ~5.5 hours.
Q1 and Q2 went the way I want all of these to go: I made a mistake (adding the marginals instead of using the addition rule properly on Q1), caught it myself because the result was mathematically impossible, and fixed it. Q2’s table I got right cold, first try.
Q3 was the low point of this whole assignment. I said, in the moment, “I feel stupid. I know I am not — I know I am smart at a lot of stuff, but I am struggling with this stats stuff!” and “I am making assumptions that are not based in anything.” Part of what made it worse was the stakes: “if I get it wrong I get it wrong there is no makeup.” That’s the closest this session came to me wanting to walk away from it — not a moment where I said I was quitting, but the moment where the frustration was most real and most visible in my own words.
I didn’t disengage, though. Instead of just grinding on Q3 in isolation, I asked for a better mental model — the cause/effect (forward likelihood vs. backward Bayes’) reframing — and once I had it, both Q3 and Q4 got noticeably easier to set up. Q4 itself was long: a lot of small, specific corrections (dropping a binomial coefficient, misremembering that \(p^0 = 1\), briefly substituting a prior probability where a per-flip probability belonged), and one exchange where my frustration came through sharply. None of it made me stop — I kept working piece by piece until all four questions were fully resolved, and then turned the whole experience into something reusable: the Claude Code skill that generates this exact report.
Spoons it took: on a 12-spoon day, I’d put this one around 8-9 spoons — this is my own read of the transcript, not a measurement. Q1/Q2 barely registered; Q3’s low point and the sheer volume of small Q4 corrections were where most of the cost actually was.
Growth rating for this session: 8/10. Evidence for it, not just a number: I went from an addition-rule error I had to be walked through on Q1, to independently asking for and applying a reframing technique (cause/effect) on Q3 and Q4 without being handed it, to catching my own errors in real time by the end (“I got 3.75 which is higher than 1…”). The two points I’m not claiming: I still needed the missing-binomial-coefficient error pointed out to me rather than catching it myself, and Q3’s table-building process took real outside help to get moving.
Skills exercised this session: the addition rule for
unions, conditional probability, joint-probability tables (row/column
reconciliation), testing for independence (joint vs. product of
marginals), the law of total probability, Bayes’ theorem in both
directions (forward likelihood vs. backward posterior), the binomial
distribution and combinatorics (\(n\)-choose-\(k\), factorials), basic R
(dbinom(), vectors, arithmetic), and — the one that isn’t a
stats skill — catching my own impossible results before writing them
down as final.
This report was generated using a custom Claude Code skill built by Dr. Teresa Vasquez. The skill takes the assignment PDF and a PDF of my own worked-out solutions — including notes on my reasoning, where I struggled, and what I learned — and digitizes that existing work into this formatted report. AI is not solving the problems; it is transcribing and formatting work I have already completed by hand. The original uploaded files (assignment PDF, handwritten work PDF, and chat transcript) are included in the submission ZIP, which is available to professors.
If 40% of DSI students access to ACCRE (Advanced Computing Center for Research and Education), 20% have access to DGX server for GPU-computing, and 8% have access to both:
Approach. My first instinct was to just add the 40% and 20% together, but once I built a small table of ACCRE / DGX / both / neither / total, that gave a probability greater than 1, which told me something was wrong. Looking at the table, I realized the 8% “both” group had already been counted once inside the 40% and once inside the 20% — so it needed to be subtracted out of each of those before I could add them, and then added back once for the union. That’s the addition rule. Once I had the “only ACCRE” and “only DGX” pieces separated out, the conditional probabilities in (b) and (c) were just a straightforward ratio of the “both” probability to the relevant marginal.
My corrected table, once I stopped double-counting the overlap:
| ACCRE only | DGX only | Both | None | Total | |
|---|---|---|---|---|---|
| P | 0.32 | 0.12 | 0.08 | 0.48 | 1.00 |
\[P(ACCRE \cup DGX) = P(ACCRE) + P(DGX) - P(ACCRE \cap DGX)\] \[P(DGX \mid ACCRE) = \frac{P(DGX \cap ACCRE)}{P(ACCRE)} \qquad P(ACCRE \mid DGX) = \frac{P(ACCRE \cap DGX)}{P(DGX)}\]
# The three numbers the problem actually hands me.
p_accre <- 0.40 # P(ACCRE) -- marginal probability of ACCRE access
p_dgx <- 0.20 # P(DGX) -- marginal probability of DGX access
p_both <- 0.08 # P(ACCRE and DGX) -- the overlap, given directly
# (a) Addition rule: add the two marginals, then subtract the overlap once so
# it isn't double-counted (it was counted inside both p_accre and p_dgx).
p_union <- p_accre + p_dgx - p_both
# (b) Conditional probability P(DGX | ACCRE) = P(DGX and ACCRE) / P(ACCRE).
# The numerator has to be the INTERSECTION (p_both), not the union -- that
# was my original mistake here.
p_dgx_given_accre <- p_both / p_accre
# (c) Same rule, flipped: P(ACCRE | DGX) = P(ACCRE and DGX) / P(DGX).
p_accre_given_dgx <- p_both / p_dgx
# Print all three so the values are visible alongside the code, not just
# quoted in the prose above.
p_union## [1] 0.52p_dgx_given_accre## [1] 0.2p_accre_given_dgx## [1] 0.4Suppose the table of probabilities described the product type and fertilizer combinations for one agricultural experiment. Compute a ~ f and:
| Product | Ammonium | Nitrogen | Total |
|---|---|---|---|
| Corn | 0.25 | a | b |
| Rice | c | d | 0.72 |
| Total | e | 0.35 | f |
Approach. This is a joint probability table, so every row has to sum to its row total and every column has to sum to its column total, and the grand total has to be 1. I worked the missing cells in the order that let each one depend only on values I already had: since the two row totals (b and 0.72) have to add to the grand total \(f = 1\), I got \(b = 1 - 0.72 = 0.28\) first. From there \(a = b - 0.25\) (row Corn), then \(d = 0.35 - a\) (column Nitrogen), then \(c = 0.72 - d\) (row Rice), and finally \(e = 0.25 + c\) (column Ammonium). For the conditional probabilities I used the general rule \(P(B \mid A) = P(B \cap A) / P(A)\) — the table already gives me all the joint and marginal values I need, so I didn’t have to convert to Bayes’ theorem. For independence, I checked whether the product of the two marginals equals the joint probability; if it doesn’t, the events are not independent.
\[P(B \mid A) = \frac{P(A \cap B)}{P(A)} \qquad P(A \cap B) = P(A) \cdot P(B) \text{ if independent}\]
# Grand total of any complete joint-probability table is always 1.
grand_total <- 1
# b = Corn's row total. Since the two row totals (b and the given 0.72 for
# Rice) must add up to the grand total, b is the one value I can get before
# anything else.
b <- grand_total - 0.72
# a = the Corn/Nitrogen cell. Corn's row is [0.25, a, b], so a is whatever's
# left after the given 0.25 is subtracted from the row total b.
a <- b - 0.25
# d = the Rice/Nitrogen cell. The Nitrogen column is [a, d, 0.35], so d is
# whatever's left after a is subtracted from the column total 0.35.
d <- 0.35 - a
# c = the Rice/Ammonium cell. Rice's row is [c, d, 0.72], so c is whatever's
# left after d is subtracted from the row total 0.72.
c <- 0.72 - d
# e = Ammonium's column total: the given 0.25 (Corn/Ammonium) plus the c I
# just solved for (Rice/Ammonium).
e <- 0.25 + c
# f = the grand total cell, always 1 for a complete table.
f <- grand_total
# (a) P(Corn | Nitrogen) = P(Corn and Nitrogen) / P(Nitrogen). The
# denominator has to be P(Nitrogen) -- the thing I'm conditioning ON -- which
# is the column total 0.35, not the row total for Corn. Dividing by the wrong
# one (0.28) was my original mistake here.
p_corn_given_nitrogen <- a / 0.35
# (b) P(Ammonium | Rice) = P(Rice and Ammonium) / P(Rice) = c / 0.72.
p_ammonium_given_rice <- c / 0.72
# (c) P(Corn and Ammonium) is a joint probability that's already sitting
# directly in the table (0.25) -- no calculation needed, and no independence
# assumption should be applied to a value that's already given.
p_corn_and_ammonium <- 0.25
# (d) Independence check: does P(Corn) * P(Ammonium) equal the actual joint
# probability P(Corn and Ammonium)? If not, the two aren't independent.
p_corn <- b
p_ammonium <- e
independence_product <- p_corn * p_ammonium
# Print the solved table cells and every answer.
c(a = a, b = b, c = c, d = d, e = e, f = f)## a b c d e f
## 0.03 0.28 0.40 0.32 0.65 1.00p_corn_given_nitrogen## [1] 0.08571429p_ammonium_given_rice## [1] 0.5555556p_corn_and_ammonium## [1] 0.25independence_product## [1] 0.182There are three machines A, B, and C generating screws with a defect rate of 0.1, 0.05, and 0.005, respectively. Machine C generates four times as many screws as B, and machine B generates three times as many screws as machine A.
Approach. The tricky part here was figuring out the units — the machines don’t produce equal shares, so I couldn’t just average the three defect rates. I set up a “units” scale using the given ratios: let A produce 1 unit, then B produces 3 units (3x A), and C produces 12 units (4x B). That gives 16 total units, so each machine’s share of production is units/16 — that’s \(P(Machine)\). From there, defectiveness is just the law of total probability: multiply each machine’s probability of being the source by its own defect rate, and sum across all three. For part (b) I already had \(P(C \cap Defective)\) from the total-probability table, and I already had \(P(Defective)\) as the sum, so I just plugged both into Bayes’ theorem instead of re-deriving it from scratch.
My production/defect table, once I had the units figured out:
| Machine | Units | P(Machine) | Defect rate | P(Machine ∩ Defective) |
|---|---|---|---|---|
| A | 1 | 1/16 ≈ 0.0625 | 0.1 | 0.00625 |
| B | 3 | 3/16 = 0.1875 | 0.05 | 0.009375 |
| C | 12 | 12/16 = 0.75 | 0.005 | 0.00375 |
| Total | 16 | 1.00 | — | 0.019375 |
\[P(Defective) = \sum_{i} P(Machine_i) \cdot P(Defective \mid Machine_i)\] \[P(C \mid Defective) = \frac{P(C \cap Defective)}{P(Defective)}\]
# The ratio language ("4x as many as B", "3x as many as A") turns into a
# units scale: pick A = 1 unit, then B and C follow from the stated ratios.
units <- c(A = 1, B = 3, C = 12)
# Each machine's share of TOTAL production is its units divided by the sum
# of all units -- this is P(Machine), the "cause" side of the problem.
p_machine <- units / sum(units)
# Given defect rates per machine -- these are P(Defective | Machine), the
# forward/likelihood direction (already handed to me, no derivation needed).
defect_rate <- c(A = 0.1, B = 0.05, C = 0.005)
# Multiplying P(Machine) by P(Defective | Machine) gives the JOINT
# probability P(Machine and Defective) for each machine.
p_machine_and_defect <- p_machine * defect_rate
# (a) Law of total probability: P(Defective) is the sum of the joint
# probabilities across all three (mutually exclusive, exhaustive) machines.
p_defect <- sum(p_machine_and_defect)
# (b) Bayes' theorem, backward direction: P(C | Defective) = P(C and
# Defective) / P(Defective). The denominator is the OVERALL P(Defective) I
# just computed -- not P(C) alone, which was my original mistake here.
p_c_given_defect <- p_machine_and_defect["C"] / p_defect
# Print the production shares, the joint probabilities, and both answers.
p_machine## A B C
## 0.0625 0.1875 0.7500p_machine_and_defect## A B C
## 0.006250 0.009375 0.003750p_defect## [1] 0.019375p_c_given_defect## C
## 0.1935484Suppose there are 10 coins, 7 of which are fair and three with \(P(tail) = 0.25\). A coin is randomly selected and flipped 5 times. Calculate the following:
Approach. For part (a), I started by writing out Bayes’ theorem across the two coin categories (fair, biased). Getting \(P(TTTTT \mid \text{coin})\) tripped me up at first — I began treating it like a binomial “choose” problem, but then realized that a specific sequence of 5 tails isn’t a combinations problem at all; it’s just the tail probability raised to the 5th power, since each flip is independent and there’s only one way to get that exact ordered sequence. So \(P(TTTTT \mid fair) = 0.5^5\) and \(P(TTTTT \mid biased) = 0.25^5\). I combined those with the prior probability of picking each coin type (7/10 fair, 3/10 biased) using the law of total probability to get the denominator, then divided to get the posterior. For part (b), “2 heads in 5 flips” is a genuine binomial question — order doesn’t matter, so this one does use the binomial coefficient \(\binom{5}{2}\), with the biased coin’s head probability \(P(head) = 1 - 0.25 = 0.75\).
My coin table, once I had both branches worked out:
| Coin | P(Heads) | P(Tails) | P(5 T’s | Coin) | P(Coin) | P(Coin) × P(5 T’s | Coin) |
|---|---|---|---|---|---|
| Fair | 0.5 | 0.5 | 0.55 = 0.03125 | 7/10 = 0.7 | 0.021875 |
| Biased | 0.75 | 0.25 | 0.255 = 0.0009766 | 3/10 = 0.3 | 0.0002930 |
| Total (P(5 T’s)) | 0.0221680 |
\[P(fair \mid TTTTT) = \frac{P(TTTTT \mid fair) \cdot P(fair)}{P(TTTTT \mid fair) \cdot P(fair) + P(TTTTT \mid biased) \cdot P(biased)}\] \[P(X = 2) = \binom{5}{2} p^2 (1-p)^3\]
# Priors: how likely I am to have picked each TYPE of coin before any flips
# happen at all -- 7 of the 10 coins are fair, 3 are biased.
p_fair_prior <- 7 / 10
p_biased_prior <- 3 / 10
# Per-flip tail probability for each coin type -- these come straight from
# the problem statement.
p_tail_fair <- 0.5
p_tail_biased <- 0.25
# P(TTTTT | coin): the probability of THIS EXACT sequence of 5 tails, given
# a coin type. Because there's only one ordering that produces "all tails",
# this collapses to just the tail probability raised to the 5th power -- no
# binomial coefficient needed here (that's only for counting orderings).
p_ttttt_given_fair <- p_tail_fair^5
p_ttttt_given_biased <- p_tail_biased^5
# Law of total probability: the OVERALL P(TTTTT), weighting each branch by
# how likely that coin type was to be picked in the first place.
p_ttttt <- p_ttttt_given_fair * p_fair_prior + p_ttttt_given_biased * p_biased_prior
# (a) Bayes' theorem, backward direction: P(fair | TTTTT) = [P(TTTTT | fair)
# times P(fair)] divided by the OVERALL P(TTTTT) computed above.
p_fair_given_ttttt <- (p_ttttt_given_fair * p_fair_prior) / p_ttttt
# (b) "2 heads in 5 flips" cares about how MANY heads, not which specific
# order they land in -- that's what makes this a real binomial question,
# unlike the single-sequence TTTTT case above. Biased coin: P(head) = 1 -
# P(tail) = 1 - 0.25 = 0.75. dbinom(x, size, prob) already includes the
# (5 choose 2) counting term internally, so it can't get dropped by accident
# the way it did in my first handwritten attempt.
p_head_biased <- 1 - 0.25
p_2heads_biased <- dbinom(2, size = 5, prob = p_head_biased)
# Print both branch likelihoods, the overall P(TTTTT), and both answers.
p_ttttt_given_fair## [1] 0.03125p_ttttt_given_biased## [1] 0.0009765625p_ttttt## [1] 0.02216797p_fair_given_ttttt## [1] 0.9867841p_2heads_biased## [1] 0.08789063