M3 - Probability models and assumptions

M3 - Probability models and assumptions

Build probability models from clearly defined events and assumptions, then check what their predictions mean. Refine dependence, population weights and physical constraints, and explain how each change affects the answer and what uncertainty remains.

Constructing a probability model

Two pumps each have a 10%10\% chance of failure during a day. Is that enough information to find the chance their system works? You also need to know how the pumps work together and how their failures are related.

A backup system works when at least one pump works. A system requiring both pumps has a different success event, even with exactly the same equipment failure probabilities.

You already know

For independent events, multiply marginal probabilities to find their intersection. In general use a conditional second probability. Complement and addition rules combine the resulting disjoint outcomes.

Before choosing any rule, define one experiment, its population and the time period. Selecting a working day uniformly differs from selecting a repair call: days with many calls contribute more observations to the second experiment.

Probability model

A probability model simplifies a chance process by specifying its possible outcomes or variables, their probabilities and relationships. It must assign nonnegative probabilities totalling 11, and its assumptions must connect those probabilities to the stated situation.

Equal likelihood supports count-to-total fractions. Independence supports products of marginal probabilities. Stable, representative rates support using past data for a future population. These assumptions are distinct: a constant individual failure rate does not establish independence between pumps.

Worked example 1: a backup arrangement

Question. Each of two pumps fails during a day with probability 0.100.10, independently. A system works if at least one pump works. Construct the four joint-outcome probabilities and find the system's operating probability.

Plan. Define the failure events, form the four disjoint outcomes, then complement total failure.

Let AA and BB denote the two failures.

P(A∩B)=0.10(0.10)=0.01both failP(A∩B′)=0.10(0.90)=0.09only A failsP(A′∩B)=0.90(0.10)=0.09only B failsP(A′∩B′)=0.90(0.90)=0.81neither failsP(system works)=1−0.01=0.99at least one works\begin{aligned} \mathrm P(A\cap B)&=0.10(0.10)=0.01 && \text{both fail} \\ \mathrm P(A\cap B')&=0.10(0.90)=0.09 && \text{only A fails} \\ \mathrm P(A'\cap B)&=0.90(0.10)=0.09 && \text{only B fails} \\ \mathrm P(A'\cap B')&=0.90(0.90)=0.81 && \text{neither fails} \\ \mathrm P(\text{system works})&=1-0.01=0.99 && \text{at least one works} \end{aligned}

Why this step? A backup arrangement fails only at the intersection of the two failure events.

Answer. The four probabilities are 0.01,0.09,0.09,0.810.01,0.09,0.09,0.81; the system works with probability 0.990.99.

Check. The four cells sum to 11, and the three working cells sum to 0.990.99.

Change only the system requirement. The failure probabilities and independence assumption stay the same.

Worked example 2: both pumps required

Question. Two pumps still fail independently with probability 0.100.10 each during a day. This system works only if both pumps work. Find its operating and failure probabilities.

Plan. Select the neither-fails cell from the same four-outcome model.

P(system works)=P(A′∩B′)both required=0.90(0.90)=0.81P(system fails)=1−0.81=0.19at least one failure\begin{aligned} \mathrm P(\text{system works})&=\mathrm P(A'\cap B') && \text{both required} \\ &=0.90(0.90)=0.81 \\ \mathrm P(\text{system fails})&=1-0.81=0.19 && \text{at least one failure} \end{aligned}

Why this step? The physical arrangement determines whether one failure is enough to stop the system.

Answer. Operating probability 0.810.81; failure probability 0.190.19.

Check. Failure is 0.01+0.09+0.09=0.190.01+0.09+0.09=0.19, the sum of the other three cells.

These reliability expressions are consequences of the defined system and probability rules; they are not separate formulae in the booklet. The addition and conditional multiplication rules are supplied; independence and complement rules must be learned.

Watch out

Wrong: “Each pump works with probability 0.90.9, so the system works with probability 0.90.9.” An individual pump is not the whole system. Define the system's success event and combine the appropriate outcomes.

A useful assumption criticism names a mechanism. For example, a shared power failure could make the two pump failures dependent.

Refining dependence assumptions

A shared cause can change simultaneous failure without changing either pump's individual failure probability. To refine the model, specify a joint or conditional probability that expresses this relationship. Saying only “dependent” does not tell you which way the answer changes.

For the backup pumps, retain P(A)=P(B)=0.10\mathrm P(A)=\mathrm P(B)=0.10. Write c=P(A∩B)c=\mathrm P(A\cap B). The four cells must then be c,0.10−c,0.10−c,0.80+cc,0.10-c,0.10-c,0.80+c. Nonnegativity requires 0≤c≤0.100\leq c\leq0.10, and the system's operating probability is 1−c1-c, between 0.900.90 and 11. Those are derived bounds, not supplied booklet formulae.

This is a striking limitation of the original information: two 90%90\%-reliable pumps can give a backup operating probability anywhere from 90%90\% to 100%100\%, depending on their relationship. Under independence it is 99%99\%; that value is one model, rather than a consequence of the two individual rates alone.

Worked example 3: more simultaneous failures

Question. For the backup system, retain P(A)=P(B)=0.10\mathrm P(A)=\mathrm P(B)=0.10, but set P(A∩B)=0.04\mathrm P(A\cap B)=0.04. Find the four joint cells, P(B∣A)\mathrm P(B\mid A), P(B∣A′)\mathrm P(B\mid A') and the system's operating probability. Compare with independence.

Plan. Subtract the overlap from each marginal, complete the table and condition on the appropriate row.

P(A∩B)=0.04given refinementP(A∩B′)=P(A′∩B)=0.10−0.04=0.06P(A′∩B′)=1−0.04−0.06−0.06=0.84P(B∣A)=0.040.10=0.4P(B∣A′)=0.060.90=115P(system works)=1−0.04=0.96\begin{aligned} \mathrm P(A\cap B)&=0.04 && \text{given refinement} \\ \mathrm P(A\cap B')&=\mathrm P(A'\cap B)=0.10-0.04=0.06 \\ \mathrm P(A'\cap B')&=1-0.04-0.06-0.06=0.84 \\ \mathrm P(B\mid A)&=\frac{0.04}{0.10}=0.4 \\ \mathrm P(B\mid A')&=\frac{0.06}{0.90}=\frac1{15} \\ \mathrm P(\text{system works})&=1-0.04=0.96 \end{aligned}

Why this step? The overlap fixes both conditional branches; retaining 0.100.10 on them would contradict the refined joint model.

Answer. Cells 0.04,0.06,0.06,0.840.04,0.06,0.06,0.84; conditionals 0.40.4 and 1/151/15; operating probability 0.960.96, down from 0.990.99.

Check. The cells sum to 11 and both failure margins remain 0.100.10.

The comparison is visible in the two complete tables: the common-failure cell grows while each failure margin stays fixed.

Independent pump model with cells 0.01, 0.09, 0.09, 0.81 and dependent model with cells 0.04, 0.06, 0.06, 0.84

Change only the overlap to 0.0050.005. Dependence can also reduce simultaneous failure.

Worked example 4: fewer simultaneous failures

Question. The same backup pumps have failure probabilities 0.100.10 each, now with P(A∩B)=0.005\mathrm P(A\cap B)=0.005. Find the four cells, the two B-failure conditional probabilities and the operating probability. Compare with independence.

Plan. Use the same cell construction with the new overlap.

P(A∩B)=0.005P(A∩B′)=P(A′∩B)=0.10−0.005=0.095P(A′∩B′)=1−0.005−0.095−0.095=0.805P(B∣A)=0.0050.10=0.05P(B∣A′)=0.0950.90=19180P(system works)=1−0.005=0.995\begin{aligned} \mathrm P(A\cap B)&=0.005 \\ \mathrm P(A\cap B')&=\mathrm P(A'\cap B)=0.10-0.005=0.095 \\ \mathrm P(A'\cap B')&=1-0.005-0.095-0.095=0.805 \\ \mathrm P(B\mid A)&=\frac{0.005}{0.10}=0.05 \\ \mathrm P(B\mid A')&=\frac{0.095}{0.90}=\frac{19}{180} \\ \mathrm P(\text{system works})&=1-0.005=0.995 \end{aligned}

Why this step? The operating event is still the complement of both failures; its change follows the specified overlap, rather than the word “dependent”.

Answer. Cells 0.005,0.095,0.095,0.8050.005,0.095,0.095,0.805; conditionals 0.050.05 and 19/18019/180; operating probability 0.9950.995, above 0.990.99.

Check. Each failure margin is 0.100.10, and 0.005<0.010.005<0.01 explains the gain over independence.

At the extremes, c=0c=0 means simultaneous failure has probability zero, while c=0.10c=0.10 means the two failure statuses agree with probability one. Neither conclusion follows merely from equal marginal rates.

Watch out

Wrong: “Dependence always lowers backup reliability.” In example 4, the overlap is below the independent value, so reliability rises. State how the refined relationship changes the particular failure event.

Before interpreting any refinement, check that its cells are valid probabilities.

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