Advanced Quantitative Concepts

Why Probability Matters in Modern Business

Every business decision involves uncertainty. Marketing teams ask: “What’s the probability this customer will convert?” Product teams ask: “What’s the probability our feature will succeed?” Operations teams ask: “What’s the probability we’ll meet our target?”

Probability quantifies uncertainty. Companies test your probability thinking because they need employees who can make data-driven decisions despite incomplete information.

Fundamental Concept: Sample Space and Events

Sample Space (S) = All possible outcomes

Event (E) = Specific outcomes we’re interested in

Probability Formula:
P(E)=Number of favorable outcomesTotal number of possible outcomes=n(E)n(S)P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{n(E)}{n(S)}P(E)=Total number of possible outcomesNumber of favorable outcomes=n(S)n(E)

Simple Example: Rolling a Die

Sample Space = {1, 2, 3, 4, 5, 6}
Total possible outcomes = 6

Event: Getting an even number = {2, 4, 6}
Favorable outcomes = 3

P(even)=36=12=0.5 or 50%P(\text{even}) = \frac{3}{6} = \frac{1}{2} = 0.5 \text{ or } 50\%P(even)=63=21=0.5 or 50%

Understanding Probability Range

• P(E) = 0 → Event is impossible
  
• 0 < P(E) < 1 → Event is possible
  
• P(E) = 1 → Event is certain
  

Critical Concept: Independent vs. Dependent Events

Two events are independent if one happening doesn’t affect the other.

Two events are dependent if one happening changes the probability of the other.

Independent Events Example:

• Flipping a coin twice
  
• Rolling two dice
  
• Drawing a card, replacing it, then drawing again
  

For independent events:
P(A and B)=P(A)×P(B)P(A \text{ and } B) = P(A) \times P(B)P(A and B)=P(A)×P(B)

Example: What’s the probability of getting heads twice in a row?

P(HH)=P(first H)×P(second H)=12×12=14=25%P(\text{HH}) = P(\text{first H}) \times P(\text{second H}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = 25\%P(HH)=P(first H)×P(second H)=21×21=41=25%

Dependent Events Example:

• Drawing 2 cards from a deck without replacement
  
• Selecting 2 people from a group without repetition
  

For dependent events:
P(A and B)=P(A)×P(B∣A)P(A \text{ and } B) = P(A) \times P(B|A)P(A and B)=P(A)×P(B∣A)

Where P(B|A) means “probability of B given A already happened”

Example: What’s the probability of drawing 2 aces from a standard deck (without replacement)?

• Total cards = 52, Total aces = 4
  
• P(first ace) = 4/52
  
• After drawing one ace: 51 cards left, 3 aces left
  
• P(second ace | first ace drawn) = 3/51
  

P(both aces)=452×351=122,652=1221≈0.45%P(\text{both aces}) = \frac{4}{52} \times \frac{3}{51} = \frac{12}{2,652} = \frac{1}{221} \approx 0.45\%P(both aces)=524×513=2,65212=2211≈0.45%

Key Difference: Notice how the second probability changed (from 4/52 to 3/51) because we removed a card. That’s dependency.

Important Rule: Addition Rule for Mutually Exclusive Events

Two events are mutually exclusive if they can’t happen simultaneously (e.g., rolling a 3 and a 5 on one die).

For mutually exclusive events:
P(A or B)=P(A)+P(B)P(A \text{ or } B) = P(A) + P(B)P(A or B)=P(A)+P(B)

Example: Probability of rolling either 3 or 5 on one die.

P(3 or 5)=16+16=26=13P(3 \text{ or } 5) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}P(3 or 5)=61+61=62=31

Important Rule: Addition Rule for Non-Exclusive Events

Sometimes events can happen together (e.g., selecting a card that’s both red AND a face card).

For non-exclusive events:
P(A or B)=P(A)+P(B)−P(A and B)P(A \text{ or } B) = P(A) + P(B) – P(A \text{ and } B)P(A or B)=P(A)+P(B)−P(A and B)

Example: From a deck, what’s the probability of drawing a red card OR a face card?

• P(red) = 26/52
  
• P(face card) = 12/52
  
• P(red AND face card) = 6/52 (red face cards)
  

P(red or face)=2652+1252−652=3252=813P(\text{red or face}) = \frac{26}{52} + \frac{12}{52} – \frac{6}{52} = \frac{32}{52} = \frac{8}{13}P(red or face)=5226+5212−526=5232=138

Complement Rule: “At Least One” Problems

Many placement problems ask: “What’s the probability of at least one successful outcome?”

Instead of calculating directly (which involves many cases), use the complement:

P(at least one)=1−P(none)P(\text{at least one}) = 1 – P(\text{none})P(at least one)=1−P(none)

Example: If 3 people each have a 60% chance of passing an exam, what’s the probability that at least one passes?

• P(one person passes) = 0.6
  
• P(one person fails) = 0.4
  
• P(all three fail) = 0.4 × 0.4 × 0.4 = 0.064
  

P(at least one passes)=1−0.064=0.936=93.6%P(\text{at least one passes}) = 1 – 0.064 = 0.936 = 93.6\%P(at least one passes)=1−0.064=0.936=93.6%

This is much faster than calculating P(exactly 1 passes) + P(exactly 2 pass) + P(all 3 pass).

Placement Test Problem: Combination + Probability

“A committee of 3 is selected from 6 men and 4 women. What’s the probability that the committee has at least 1 woman?”

Total ways to select 3 from 10: 10C3 = 120

Method 1: Direct Calculation

• 1 woman, 2 men: 4C1 × 6C2 = 4 × 15 = 60
  
• 2 women, 1 man: 4C2 × 6C1 = 6 × 6 = 36
  
• 3 women, 0 men: 4C3 × 6C0 = 4 × 1 = 4
  
• Total favorable = 60 + 36 + 4 = 100
  

P(at least 1 woman)=100120=56P(\text{at least 1 woman}) = \frac{100}{120} = \frac{5}{6}P(at least 1 woman)=120100=65

Method 2: Complement (Faster)

• No women (all 3 men): 6C3 = 20
  
• P(all men) = 20/120 = 1/6
  

P(at least 1 woman)=1−16=56P(\text{at least 1 woman}) = 1 – \frac{1}{6} = \frac{5}{6}P(at least 1 woman)=1−61=65

Notice how the complement method is cleaner and faster.