# Q1(a) Joint masses, normalization and conditional probability [5]

Subject: Intelligent Systems II

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-2025-april/q1a

Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-2025-april--q1a/

Actual historical paper ·100 marks attempted /140 offered · Q1 compulsory + any four of Q2–Q7. Source defects and solution assumptions are explicitly identified.

**Question. Q1(a)** [5 marks · 2025-04-28]

| Machine | No Fault (X = 0) | Fault (X = 1) |
| --- | --- | --- |
| Machine A | 0.40 | 0.05 |
| Machine B | 0.25 | 0.10 |
| Machine C | 0.20 | 0.08 |
| Machine D | 0.05 | 0.07 |

A factory uses four machines (Machine A, Machine B, Machine C, and Machine D) for production. The above table shows the joint probability distribution of a fault occurring (X) and the machine type (Y).

- If a fault occurs, what is the probability that it came from Machine C?
- What is the probability that a randomly selected product has a fault?
- Given the product is from Machine B, what is the probability that it has a fault?

### Answer

### Given

The eight printed entries are used exactly as shown. X=1 means fault; Y identifies the machine.

### To find

P(C | fault), P(fault), and P(fault | B).

### Formula and why



$$
P(A\mid B)=\frac{P(A\cap B)}{P(B)}
$$



Conditioning restricts the denominator to the given group. Before treating a whole table as a joint probability distribution, its entries must sum to 1.

### Step 1. Check the printed total



$$
\text{No-fault total}=0.40+0.25+0.20+0.05=0.90
$$





$$
\text{Fault total}=0.05+0.10+0.08+0.07=0.30
$$





$$
\text{Grand total}=0.90+0.30=1.20\ne1
$$



**Source inconsistency:** as printed, this is not a valid normalized probability table. Treating the entries as proportional nonnegative masses is one explicit repair; dividing every entry by 1.20 preserves conditional ratios.

### Step 2. Machine C, given a fault



$$
P(C\mid X=1)=\frac{0.08/1.20}{0.30/1.20}=\frac{0.08}{0.30}=\frac4{15}\approx0.266667
$$



Among all fault mass, 0.08 belongs to C. The common normalization factor cancels.

### Step 3. Overall fault probability under the stated repair



$$
P(X=1)=\frac{0.30}{1.20}=0.25
$$



**Answer under mass normalization: 25%.** The raw column sum 0.30 is only the unnormalized fault mass. If the examiner intended another corrected entry, the unconditional probability depends on that correction; 0.30 cannot be asserted as a valid probability while retaining all eight printed numbers.

### Step 4. Fault, given Machine B



$$
P(X=1\mid B)=\frac{0.10}{0.25+0.10}=\frac{0.10}{0.35}=\frac27\approx0.285714
$$



Machine B is the denominator because the product is known to come from B. **Final conditional answers: 26.6667% and 28.5714%.**