🔬 Session 2.3 - Understanding Probability with Real Examples

Example 2.12 – Laser Diodes

We have 100 laser diodes. 30 meet the customer's requirement. Green = pass.

Probability of selecting a good diode (Event E): P(E) = 30 / 100 = 0.30

Example 2.13 – Events and Probabilities with step-by-step explanations

Sample space: S = {a, b, c, d}

Assigned probabilities: P(a)=0.1, P(b)=0.3, P(c)=0.5, P(d)=0.1

Events: A = {a, b}, B = {b, c, d}, C = {d}

S = {a,b,c,d} | A = {a,b} | A′ = {c,d}
B = {b,c,d} | B′ = {a}
C = {d} | C′ = {a,b,c}

P(A) = 0.1 + 0.3 = 0.4
Why?

A = {a,b}. Add the probabilities of the outcomes inside A: P(a)=0.1 and P(b)=0.3. Because outcomes in S are disjoint and their probabilities add, P(A)=P({a})+P({b})=0.1+0.3=0.4.
P(B) = 0.3 + 0.5 + 0.1 = 0.9
Why?

B = {b,c,d}. Sum the masses on b, c, d: P(b)=0.3, P(c)=0.5, P(d)=0.1. So P(B)=0.9.
P(C) = 0.1
Why?

C = {d}. Single outcome d has probability 0.1, therefore P(C)=0.1.
P(A′) = 1 − 0.4 = 0.6
Why?

The complement A′ contains everything not in A. From the set readout above, A′={c,d}. In probability, P(A′)=1−P(A) because A and A′ partition S. Since P(A)=0.4, we get 0.6. (You can also check: P(c)+P(d)=0.5+0.1=0.6.)
P(B′) = 1 − 0.9 = 0.1
Why?

B′ is everything not in B. From sets: B′={a}. Using the complement rule, P(B′)=1−P(B)=1−0.9=0.1. (Matches P(a)=0.1.)
P(C′) = 1 − 0.1 = 0.9
Why?

C={d} so C′={a,b,c}. Complement rule gives 1−P(C)=0.9. Check by addition: P(a)+P(b)+P(c)=0.1+0.3+0.5=0.9.
P(A ∩ B) = P({b}) = 0.3
Why?

A∩B means outcomes that are in both A and B. With A={a,b} and B={b,c,d}, the only common element is {b}. Therefore P(A∩B)=P(b)=0.3.
P(A ∪ B) = 1 (full set)
Why?

A∪B contains anything in A or in B (or both). Here, A∪B = {a,b} ∪ {b,c,d} = {a,b,c,d} = S. The probability of the whole sample space is 1, so P(A∪B)=1.

Alternate check with inclusion–exclusion: P(A)+P(B)−P(A∩B)=0.4+0.9−0.3=1.0.
P(A ∩ C) = 0 (null set)
Why?

A={a,b} and C={d} have no common outcomes, so A∩C=∅. The probability of the empty set is 0: P(∅)=0.

Example 2.14 – Manufacturing Inspection (Hypergeometric)

Real-World Setup: A bin has 50 parts: 3 defective, 47 good. Select 6 without replacement.

Goal: Probability that exactly 2 of the 6 parts are defective.

Step-by-step breakdown:

Choose 2 defective from 3: C(3,2)
Choose 4 good from 47: C(47,4)
Favorable samples: C(3,2) × C(47,4) = 3 × 178,365 = 535,095
Total samples: C(50,6) = 15,890,700
Probability = 535,095 / 15,890,700 ≈ 0.034

Another case: 0 defectives (all good)?

Pick all 6 from 47 good: C(47,6) = 10,737,573
Probability = 10,737,573 / 15,890,700 ≈ 0.676

Takeaway: With only 3 defectives in 50, it’s much more likely your 6 are all good. This is the hypergeometric setting (sampling without replacement).

📏 Axioms of Probability

P(S) = 1 (Sample space has total probability 1)
0 ≤ P(E) ≤ 1 for any event E
If E₁ and E₂ are disjoint: P(E₁ ∪ E₂) = P(E₁) + P(E₂)

Other useful rules:

P(∅) = 0
P(E′) = 1 − P(E)
If E₁ ⊆ E₂, then P(E₁) ≤ P(E₂)

FYI – Advanced Rules:

Inclusion–Exclusion: P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Complement of union: P((A ∪ B)′) = 1 − P(A ∪ B)
Mutually exclusive A₁,…,Aₙ: P(A₁ ∪ … ∪ Aₙ) = Σ P(Aᵢ)