MATH U113 · Doubt clinic · topic page
Independence, reliability diagrams, repeated trials
Independence is a computation; it does not survive conditioning on a system built from the event; pairwise is not mutual; sequences multiply within and add across. Notes: §2.5. Cards are collapsed — open one, or use the buttons. Newest at the bottom. Back to the clinic index.
Problem set Q9 · 14 SepFour-component system; C fails given it worksA and B and (C or D): P(system works), and P(C fails | system works) — why it is not 0.2.
Where this lives Notes §2.5 · reliability, worked example 3 · Notes §2.4 · conditional probability · Devore §2.5
(a) is the series–parallel recipe from the notes. (b) is where people stop: "the components are independent, so P(C fails | works) = 0.2"? No — C is independent of A, B, D, but not of "the system works", which is built partly from C. Use the definition of conditional probability.
Step 1Draw the block diagram, reduce it
Step 2(b) by the definition, with the numerator built from the components
"C fails and the system works" forces a specific pattern: A works, B works, C fails, and therefore D must work. Independence lets us multiply those four:
Notice the 0.81 cancels: the answer is (0.2 × 0.8)/0.96 — only the parallel block matters, because A and B working is required in both numerator and denominator. Sanity: 1/6 < 0.2 ✓ — knowing the system works makes "C failed" less likely.
1) Answering 0.2 for (b). Independence of components does not make C independent of the system's state. 2) Numerator = P(Cc) · P(W). Same error in formula form. 3) Parallel as addition. P(C or D) = 0.8 + 0.8 − 0.64 = 0.96, or 1 − 0.04 — never 1.6.
Try the twin: P(both C and D work | system works)?
P(C ∩ D ∩ W) = 0.81 × 0.8 × 0.8 = 0.5184; divide by 0.7776: 0.64/0.96 = 2/3 ≈ 0.66667. The three cases given W — both work (2/3), only D (1/6), only C (1/6) — add to 1 ✓.
Problem set Q10 · 14 SepPairwise vs mutual independence on two diceRed 3, green 4, total 7: pairwise independent (yes) but not mutually (no) — checked on the grid.
Where this lives Notes §2.5 · independence; pairwise vs mutual (new paragraph) · Doubt card D2 (the 36-cell grid) · Devore §2.5
Independence is a computation, not a feeling: check whether P(A ∩ B) equals P(A)P(B). "Pairwise" means the check passes for each of the three pairs; "mutually" additionally needs P(A ∩ B ∩ C) = P(A)P(B)P(C). This question is the standard example where the first holds and the second fails.
Step 1Each event on the 36-cell grid
A: the row red = 3, six cells, P(A) = 1/6. B: the column green = 4, six cells, 1/6. C: the anti-diagonal of sums 7 — (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) — six cells, 1/6.
Step 2(a) The three pairwise checks
Each pair intersects in the single cell (3, 4): P(A ∩ B) = P(A ∩ C) = P(B ∩ C) = 1/36. And each product is (1/6)(1/6) = 1/36. All three checks pass — pairwise independent: yes. (For A, C in words: knowing the red die is 3, the total is 7 exactly when green is 4 — still 1/6.)
Step 3(b) The triple check
Not equal — mutually independent: no. In words: knowing A and B (red 3, green 4) fixes the total at 7, so C becomes certain: P(C | A ∩ B) = 1 ≠ 1/6. Two pieces of information together tell you what neither told you alone.
1) Pairwise ⇒ mutual. This question exists to kill that belief. Mutual independence needs every sub-collection's product rule, including the triple. 2) Judging by intuition ("C depends on both dice so it can't be independent of A"). Compute; it is. 3) Checking only the triple. The triple failing does not by itself say anything about the pairs — both checks are asked.
Try the twin: A = red even, B = green even, C = total even. Pairwise? Mutually?
Each has probability 1/2. A ∩ B: 9 cells = 1/4 = (1/2)(1/2) ✓. A ∩ C: red even and total even forces green even — 9 cells, 1/4 ✓. Same for B ∩ C ✓. Pairwise: yes. Triple: A ∩ B ∩ C = A ∩ B (both even makes the total even automatically) = 1/4 ≠ 1/8. Mutually: no.
Problem set Q12 · 14 SepThe circuit: operates, and which device is faultyThree parallel branches with failure probabilities: P(operates), P(operates | C or E fails), P(C or E faulty | dead).
Where this lives Notes §2.5 · reliability · Notes §2.4 · Bayes · Tutorial 2 Q5 from zero (Bayes with a built likelihood) · Devore §2.4–2.5
Three separate skills stacked: read the topology (three parallel branches, one of which contains a parallel pair in series with E); work with failure probabilities (the diagram gives failures, so "works" = 1 − number); and then, in (d)–(e), run Bayes with "does not operate" as the evidence. Do the reduction once, carefully, and (b)–(e) are re-runs with one device forced to fail.
Step 1Redraw and reduce, branch by branch
- Top (A, B in series): works 0.9 × 0.8 = 0.72, fails 0.28.
- Middle: the pair C ∥ D fails only if both fail, 0.1 × 0.4 = 0.04, so the pair works 0.96; in series with E (works 0.5): branch works 0.96 × 0.5 = 0.48, fails 0.52.
- Bottom (F alone): works 0.7, fails 0.3.
Three branches in parallel: the circuit fails only if all three fail.
Step 2(b), (c): re-run with one device dead
(b) C fails. The pair now works only if D works (0.6); middle branch works 0.6 × 0.5 = 0.3, fails 0.7. P(not | Cc) = 0.28 × 0.7 × 0.3 = 0.0588, so P(operates | C fails) = 0.9412.
(c) E fails. The middle branch is dead regardless of C, D: fails with probability 1. P(not | Ec) = 0.28 × 1 × 0.3 = 0.084, so P(operates | E fails) = 0.916.
Both are conditional probabilities computed by rebuilding the circuit — legitimate because the other devices are independent of the one you fixed.
Step 3(d), (e): Bayes with "does not operate" as the evidence
The likelihoods are exactly the (b) and (c) failure numbers — that is why the question asks them first. Sanity: a dead circuit almost certainly has E dead (0.96), because the middle branch can only fail through E or through both C and D, and the latter is rare (0.04). C being faulty barely matters (0.13 ≈ its prior 0.1, nudged up).
1) Treating the diagram's numbers as "works". Here they are failure probabilities; a device "works" with 1 − that. Read the caption of every diagram. 2) Middle branch as (C ∥ D) ∥ E. E is in series after the pair — the wire from the pair goes into E. 3) (d) as P(not | C fails). That is (b)'s complement, 0.0588; (d) flips the conditioning — Bayes. 4) Forgetting P(not) is tiny. Dividing by 0.04368 is what turns 0.042 into 0.96.
Try the twin: P(F faulty | circuit does not operate)?
P(not | F fails) = 0.28 × 0.52 × 1 = 0.1456; Bayes: 0.3 × 0.1456 / 0.04368 = 0.04368/0.04368 = 1. Exactly 1 — and it must be: F is a whole branch by itself, so the circuit cannot be dead unless F is. A posterior of exactly 1 is the structure talking, not a coincidence.
Problem set Q15 · 14 SepThree vehicles, 70% passAll pass, at least one fails, exactly one passes, at most one, all three given at least one — eight sequences.
Where this lives Notes §2.5 · repeated independent trials, the sequence trick · Tutorial 2 Q5 from zero (same trick) · Devore §2.5 (and §3.4 binomial, later)
All five parts are the same machine: independent trials with p = 0.7 pass, 0.3 fail. Multiply within a sequence, add across sequences. Eight sequences in total; write them once and every part is a sum of some of them.
Step 1The eight sequences and their probabilities
One sequence with k passes has probability 0.7k · 0.33−k; there are C(3, k) such sequences. (This is the binomial pmf of Module 2 arriving early — you don't need its name here.)
Step 2Read off each part
- (a) 0.73 = 0.343.
- (b) at least one fails = not all pass: 1 − 0.343 = 0.657.
- (c) exactly one passes: three sequences, 3 × 0.7 × 0.32 = 3 × 0.063 = 0.189.
- (d) at most one passes = zero or one: 0.027 + 0.189 = 0.216.
- (e) conditional: P(all three | at least one) = P(all three)/P(at least one), i.e. 0.343/(1 − 0.027) = 0.343/0.973 ≈ 0.35252 — the numerator is P(all three ∩ at least one) = P(all three), since "all three" is inside "at least one".
Check: the four column totals 0.343 + 0.441 + 0.189 + 0.027 = 1 ✓.
1) (c) as 0.7 × 0.3² = 0.063. That is one sequence; the pass can be in any of three positions. 2) (b) as 0.3 (or 3 × 0.3). "At least one fails" is a complement, not a single trial. 3) (e) with the wrong numerator. The intersection of "all pass" with "at least one passes" is "all pass" — don't multiply the two.
Try the twin: p = 0.9. Find (c) and (e).
(c) 3 × 0.9 × 0.12 = 0.027. (e) 0.729/(1 − 0.001) = 0.729/0.999 ≈ 0.72973.
Devore Example 2.35 · 15 SepTwo suppliers' batches: why the answer is three numbers multipliedOn a random day, P(two batches arrive and both pass) = (.8)(.9)(.4) = .288 — what the tree is doing.
Where this lives Notes §2.5 · independence · Notes §2.4 · multiplication rule · Strike and bags slides (the tree) · Devore §2.5
Three separate facts are being multiplied, and it is not obvious why they combine or where the .4 comes from. Decode the sentence: "two batches pass" requires two batches to have arrived (only on 2 of 5 days, so .4) and both to pass. The first "and" is the multiplication rule; the second is independence. The tree makes the two different "ands" visible.
Step 1The event has a hidden first condition
Let T = "two batches received today". P(T) = 2/5 = .4 — a randomly selected weekday is one of the two supplier-2 days with that probability. On the other three days only one batch arrives and "two batches pass" is impossible. So
Step 2Given two batches, independence multiplies the pass rates
On a two-batch day the assumption says the first batch's result does not affect the second's: P(both pass | T) = (.8)(.9) = .72. Hence
1) (.8)(.9) = .72. That is the probability given a two-batch day; the question says "on a randomly selected day", so the .4 must be there. 2) Adding .8 + .9. "Both pass" is an and. 3) Multiplying without independence. The .72 step is only legal because the book assumes independence on two-batch days; the question says so explicitly, which is your cue to use it.
Try the twin: on a random day, P(at least one batch fails inspection)?
One-batch days: fails with .2 → contributes (.6)(.2) = .12. Two-batch days: at least one fails = 1 − both pass = 1 − .72 = .28 → contributes (.4)(.28) = .112. Total .232. Check by complement: P(everything that arrived passed) = (.6)(.8) + (.4)(.72) = .48 + .288 = .768, and 1 − .768 = .232 ✓.
Devore Example 2.36 · 15 SepSolar arrays: series-parallel versus total-cross-tiedSix cells at .9 each. Two chains of three in parallel: .927. Cross-ties at every column: .970 — and why the ties help.
Where this lives Notes §2.5 · reliability, worked example 3 · Four-component system card · Circuit card · Devore §2.5
Two things. First, the book gives the series-parallel answer twice — once by the addition rule, once by "1 − both subsystems fail" — and it is not obvious they are the same calculation. Second, the cross-tied diagram looks like the same six cells, yet the answer goes up; where does the extra reliability come from? Both are answered by reducing the diagram block by block, which is the only method you need for any of these.
Step 1Series-parallel: reduce the chains, then the parallel pair
- A chain (three in series) works only if all three work: .9 × .9 × .9 = .729 (independence).
- Two chains in parallel: the system works if at least one chain works. Complement route: 1 − P(both chains fail) = 1 − (1 − .729)2 = 1 − (.271)2 = 1 − .0734 = .927.
Why the book's first calculation is the same thing. The addition rule for "chain 1 works or chain 2 works": .729 + .729 − P(both work) = .729 + .729 − (.9)6 = 1.458 − .531 = .927. Two routes, one number: the addition rule subtracts the overlap; the complement route never counts it twice in the first place. Use whichever you can write faster — for "at least one of several", the complement is almost always faster.
Step 2Total-cross-tied: the ties change the topology, not the cells
With a tie after every column, current can hop between the rows at each junction. So the system needs, in each column, at least one of the two cells working — three parallel pairs, in series:
- A column (two in parallel) fails only if both cells fail: works with 1 − (.1)(.1) = .99.
- Three columns in series: .993 = .970.
Why .970 > .927. In (a), cell 1 failing kills the whole top chain, so cells 2 and 3 become useless. In (b), cell 1 failing only forces the current through cell 4 at that column; cells 2 and 3 still help. The ties let the good cells of one row cover for the bad cells of the other, column by column. Same six cells, same .9 each — more ways to succeed.
1) Parallel = add. .729 + .729 = 1.458 is not a probability; the overlap must be subtracted, or use 1 − Π(fails). 2) Reducing (b) as two chains. The ties make it three columns in series, not two rows in parallel; read the wiring, not the picture's rows. 3) .9⁶ for (a). That is "all six work", which is only one way for the system to work. 4) Series and parallel swapped. Series: multiply the works. Parallel: multiply the fails and subtract from 1.
Try the twin: same two layouts with P(Aᵢ) = .8. Series-parallel and cross-tied reliabilities?
Chain: .8³ = .512; two in parallel: 1 − (.488)2 = 1 − .238 = .762. Column: 1 − .2² = .96; three in series: .963 = .885. The gap widens (.762 vs .885) when cells are less reliable — cross-ties matter more the weaker the parts.