Week 5 & 6 - Probability

🎯 Basic Probability

Probability of an event:
$P(E)=\frac{\text{favorable outcomes}}{\text{possible outcomes}}$

• $0\le P(E)\le 1$
• 0 → impossible, 1 → certain event

Examples

• Coin → Tails: $1/2$
• Die → 2: $1/6$
• Deck → King: $4/52$

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💡 Meaning of Probability Values

Probability	Meaning	Example
$P(E)=0$	Impossible event	Drawing a Joker from deck w/o Jokers
$0<P(E)<1$	Possible, not certain	Drawing a King from standard deck
$P(E)=1$	Certain event	Drawing a red ball when all are red

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🔗 Complex Events

Types

1. Conjunctive (A and B) — both events occur.
2. Disjunctive (A or B) — at least one event occurs.

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✖️ Conjunction Rules

Independent Events

$P(A\text{ and }B)=P(A)\times P(B)$

Examples

• Two dice (snake eyes): $(1/6)\times (1/6)=1/36$
• Two heads (2 coin tosses): $(1/2)\times (1/2)=1/4$

Dependent Events

When outcome of B depends on A:

General Multiplication Rule:
$P(A\text{ and }B)=P(A)\times P(B|A)$

Example (Drawing Kings)

• With replacement: $(4/52)\times (4/52)=16/2704$
• Without replacement: $(4/52)\times (3/51)=12/2652$

So probability decreases without replacement.

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🧮 Example — Urn Problem

Urn contains 10 balls: 3 Red, 2 Green, 5 Blue

Event	With Replacement	Without Replacement
2 Red	$(3/10)\times (3/10)=9/100$	$(3/10)\times (2/9)=6/90$
Red then Green	$(3/10)\times (2/10)=6/100$	$(3/10)\times (2/9)=6/90$

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➕ Disjunction Rules

Mutually Exclusive

$P(A\text{ or }B)=P(A)+P(B)$
→ Example: King or Queen → $(4/52)+(4/52)=8/52$

Not Mutually Exclusive

$P(A\text{ or }B)=P(A)+P(B)-P(A\text{ and }B)$
→ Avoids double‑counting overlap events.

Example (At least one Head in 2 tosses):
$1/2+1/2-1/4=3/4$

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⚖️ Mutually Exclusive Events

• Can’t both occur.
• Restricted addition rule applies: $P(AorB)=P(A)+P(B)$
• For non‑exclusive events, use the general addition rule.

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💭 Practice — Example

Urn: 2 Red, 2 Green, 6 Blue

Situation	Formula	Result
At least 1 Red (with replacement)	$2/10+2/10-(2/10\times 2/10)$	$36/100$
At least 1 Red (without replacement)	$2/10+2/10-(2/10\times 1/9)$	$34/90$
Red₁ or Blue₂ (with replacement)	$2/10+5/10-(2/10\times 5/10)$	$6/10$

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♻️ The Complementation Rule

Idea: Probabilities of all events in a sample space sum to 1.

If $A$ is an event, and $\sim A$ its complement:
$P(A)+P(\sim A)=1$
→ Therefore $P(A)=1-P(\sim A)$

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🔹 Examples

Event	Complement Form	Result
Coin → Head	$1-P(Tails)=1-1/2=1/2$
Die → Six	$1-P(\text{not six})=1-5/6=1/6$
Deck → King	$1-P(\text{not King})=1-48/52=4/52$
Urn: 3 Red/2 Green → Red	$1-2/5=3/5$

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🧩 Using Complementation for “At Least One” Problems

To find probability of at least one X, compute its complement — no Xs.

Example 1 – Two draws (3 red, 2 green, with replacement)

$P(\text{at least 1 red})=1-P(\text{no reds})$
$=1-(2/5\times 2/5)=1-4/25=21/25$

Example 2 – Without replacement

$1-(2/5\times 1/4)=1-1/10=9/10$

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📊 Extending to 3 Events

Conjunction (All happen)

• Independent: $P(A\text{ and }B\text{ and }C)=P(A)\times P(B)\times P(C)$
• Dependent: $P(A)\times P(B|A)\times P(C|A,B)$
  • e.g. 3 Kings (no replacement): $(4/52)\times (3/51)\times (2/50)$

Disjunction (At least one)

• Mutually exclusive: $P(A)+P(B)+P(C)$
• Not mutually exclusive (via complement):
  $1-P(\sim A\text{ and }\sim B\text{ and }\sim C)$

At least 1 King (3 draws, with replacement):
$1-(48/52{)}^3=0.22$

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🧩 Key Recap of Probability Rules

Type	Independent	Dependent	Mutually Exclusive	Not Mutually Exclusive
A and B	$P(A)P(B)$	$P(A)P(B	A)$	—
A or B	—	—	$P(A)+P(B)$	$P(A)+P(B)-P(A\text{and}B)$
Complement	$P(A)=1-P(\sim A)$	—	—	—
3 Events (Independent)	$P(A)P(B)P(C)$	$P(A)P(B	A)P(C	A,B)$

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🧮 Practice Quickies

1. Two Kings (no replacement): $(4/52)\times (3/51)$
2. At least one Ace (3 draws w/) $1 –(48/52{)}^3$
3. Exactly 1 Red (2 draws no replacement): $(4/10\times 6/9)+(6/10\times 4/9)=48/90$

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✅ Summary

• Complement Rule: efficient for “at least one” situations.
• Multiplication Rules: for and cases (independent vs. dependent).
• Addition Rules: for or cases (exclusive vs. overlapping).
• Sound probability reasoning: combine these via conditions & structure.