Probability Models in Engineering: Step-by-Step Guide to ECE2191 Example 1

Introduction to Probability Models in Engineering

Probability models are the backbone of decision-making in engineering, from reliability analysis to signal processing. In this tutorial, we break down ECE2191 Example 1 into digestible steps, using current trends like AI-driven predictions and school life examples to make concepts stick. Whether you're preparing for exams or building real-world systems, understanding probability mass functions (PMF), expectation, variance, and transformations is crucial.

Problem 1: Matching Gloves – Hypergeometric Distribution

Scenario: A drawer has 10 pairs of matching gloves (20 gloves total). Six gloves are selected randomly. Let X be the number of complete pairs obtained. This is a classic hypergeometric problem.

Step 1: Define the random variable

X can be 0, 1, 2, or 3 (since you can't get more than 3 pairs from 6 gloves).

Step 2: Count favorable outcomes

Total ways to choose 6 gloves from 20: C(20,6). For X = k, we need to choose k pairs (complete) and then from the remaining 20-2k gloves, choose the other 6-2k gloves such that no additional pair is formed. This is a standard hypergeometric PMF:

P(X = k) = [C(10, k) * C(10 - k, 6 - 2k) * 2^(6-2k)] / C(20, 6), for k = 0,1,2,3

Where the factor 2^(6-2k) accounts for choosing left or right glove from each of the remaining pairs.

Step 3: Compute numerically

For k=0: C(10,0)*C(10,6)*2^6 / C(20,6) = 1*210*64 / 38760 ≈ 0.346
For k=1: C(10,1)*C(9,4)*2^4 / 38760 = 10*126*16 / 38760 ≈ 0.520
For k=2: C(10,2)*C(8,2)*2^2 / 38760 = 45*28*4 / 38760 ≈ 0.130
For k=3: C(10,3)*C(7,0)*2^0 / 38760 = 120*1*1 / 38760 ≈ 0.003

This PMF sums to 1, as expected.

Problem 2: Urn Game – Expectation and Variance of Net Gain

Scenario: An urn has 5 balls: two $1, two $5, one $15. Player pays $10, draws two balls without replacement, wins the sum. Net gain = sum - $10.

Step 1: List all possible sums

Possible sums: 1+1=2, 1+5=6, 1+15=16, 5+5=10, 5+15=20, 15+15=30 (but only one $15, so 30 impossible). Actually, with replacement? No, without replacement. So possible sums: 2, 6, 16, 10, 20. Probabilities:

• P(sum=2): C(2,2)/C(5,2)=1/10
• P(sum=6): (C(2,1)*C(2,1))/10 = 4/10
• P(sum=16): (C(2,1)*C(1,1))/10 = 2/10
• P(sum=10): C(2,2)/10 = 1/10
• P(sum=20): (C(2,1)*C(1,1))/10 = 2/10

Step 2: Net gain distribution

Net gain G = sum - 10. So values: -8, -4, 6, 0, 10 with same probabilities.

Step 3: Expectation

E[G] = (-8)*0.1 + (-4)*0.4 + 6*0.2 + 0*0.1 + 10*0.2 = -0.8 -1.6 +1.2 +0 +2 = 0.8

So expected net gain is $0.80 (positive, but small).

Step 4: Variance

E[G^2] = 64*0.1 + 16*0.4 + 36*0.2 + 0*0.1 + 100*0.2 = 6.4 + 6.4 + 7.2 + 0 + 20 = 40
Var(G) = E[G^2] - (E[G])^2 = 40 - 0.64 = 39.36

High variance indicates risk.

Problem 3: Joint PMF – Finding Constant and Marginals

Scenario: Given joint PMF of X and Y (table missing in excerpt, but typical). We'll assume a common form: P(X=x, Y=y) = k * (x+y) for x=1,2; y=1,2,3.

Step 1: Find k

Sum over all x,y: k[ (1+1)+(1+2)+(1+3)+(2+1)+(2+2)+(2+3) ] = k[2+3+4+3+4+5] = k*21 = 1 => k=1/21.

Step 2: Marginal PMFs

P(X=1) = sum_y P(1,y) = (2+3+4)/21 = 9/21 = 3/7
P(X=2) = (3+4+5)/21 = 12/21 = 4/7
P(Y=1) = (2+3)/21 = 5/21
P(Y=2) = (3+4)/21 = 7/21 = 1/3
P(Y=3) = (4+5)/21 = 9/21 = 3/7

Step 3: Conditional probability

P(X>1 | Y=1) = P(X=2, Y=1) / P(Y=1) = (3/21) / (5/21) = 3/5.

Step 4: Expectations

E[X] = 1*(3/7) + 2*(4/7) = 11/7 ≈ 1.571
E[Y] = 1*(5/21) + 2*(7/21) + 3*(9/21) = (5+14+27)/21 = 46/21 ≈ 2.190

Problem 4: Transformation of Random Variable – Y = 1/X

Scenario: X ~ Uniform(0,1). Find PDF of Y = 1/X.

Step 1: CDF method

For y > 1, F_Y(y) = P(Y ≤ y) = P(1/X ≤ y) = P(X ≥ 1/y) = 1 - F_X(1/y) = 1 - 1/y.

Step 2: Differentiate

f_Y(y) = d/dy (1 - 1/y) = 1/y^2 for y > 1, and 0 otherwise.

This is a Pareto distribution (Type I).

Problem 5: Geometric Probability – Line Segment

Scenario: A point is selected uniformly on a line segment of length L. Find probability that none of the two resulting segments is smaller than L/3.

Step 1: Define the condition

Let the point be at distance x from left end. Then segments lengths are x and L-x. Condition: x ≥ L/3 and L-x ≥ L/3 => x ∈ [L/3, 2L/3].

Step 2: Probability

Length of favorable interval = L/3. Total length = L. So probability = (L/3)/L = 1/3.

Problem 6: Binomial Approximation – School Tax Vote

Scenario: 65% in favor. Sample of 100. Approximate probabilities using normal approximation (since n large).

Let X ~ Binomial(100, 0.65). Mean μ = 65, variance σ^2 = 100*0.65*0.35 = 22.75, σ ≈ 4.77.

Part 1: At least 50

P(X ≥ 50) = P(Z ≥ (49.5 - 65)/4.77) = P(Z ≥ -3.25) ≈ 0.9994 (using continuity correction).

Part 2: Between 60 and 70 inclusive

P(59.5 < X < 70.5) = P( (59.5-65)/4.77 < Z < (70.5-65)/4.77 ) = P(-1.15 < Z < 1.15) ≈ 0.75.

Part 3: Fewer than 75

P(X < 75) = P(Z < (74.5-65)/4.77) = P(Z < 1.99) ≈ 0.9767.

Connecting Probability to Current Trends

Probability models are used in AI for predictive analytics (e.g., recommending videos or products), in finance for risk assessment (like the urn game variance), and even in gaming loot boxes (hypergeometric distribution). Understanding these fundamentals helps engineers design reliable systems.

Conclusion

We've covered hypergeometric, expectation, joint PMF, transformation, geometric probability, and binomial approximation. Practice these steps to ace ECE2191 and apply probability in engineering contexts.