Episode 8 — Aptitude and Reasoning / 8.15 — Probability
8.15.a Concepts and Formulas -- Probability
1. Basic Definitions
Experiment
An activity or process that produces a definite outcome. Examples: tossing a coin, rolling a die, drawing a card.
Random Experiment
An experiment whose outcome cannot be predicted with certainty in advance.
Sample Space (S)
The set of all possible outcomes of a random experiment.
Coin toss: S = {Head, Tail} |S| = 2
Rolling a die: S = {1, 2, 3, 4, 5, 6} |S| = 6
Two coins: S = {HH, HT, TH, TT} |S| = 4
Two dice: |S| = 36
n coins: |S| = 2^n
Event (E)
A subset of the sample space. An event is said to "occur" when the outcome belongs to that subset.
Event: "getting an even number on a die"
E = {2, 4, 6}
Favourable Outcomes
The outcomes in the sample space that satisfy the condition of the event.
2. Definition of Probability
Classical (Theoretical) Definition
P(E) = Number of favourable outcomes / Total number of outcomes
= n(E) / n(S)
Properties
1. 0 <= P(E) <= 1 for any event E
2. P(S) = 1 (certain event)
3. P(empty set) = 0 (impossible event)
4. P(E) = 0 means E is impossible
5. P(E) = 1 means E is certain
Example
What is the probability of getting a prime number when a die is rolled?
S = {1, 2, 3, 4, 5, 6}
E (prime) = {2, 3, 5}
P(E) = 3/6 = 1/2
3. Complementary Probability
Definition
The complement of event E (written as E' or E-bar) consists of all outcomes NOT in E.
P(E') = 1 - P(E)
P(not E) = 1 - P(E)
Why It Matters
For "at least one" problems, it is almost always easier to compute:
P(at least one) = 1 - P(none)
Example
Probability of getting at least one head in 3 coin tosses:
P(at least 1 head) = 1 - P(no head)
= 1 - P(all tails)
= 1 - (1/2)^3
= 1 - 1/8
= 7/8
4. Types of Events
Mutually Exclusive Events
Two events A and B are mutually exclusive if they cannot occur simultaneously.
A intersection B = empty set
P(A and B) = 0
Example: Getting a "2" and getting a "5" on a single die roll are mutually exclusive.
Exhaustive Events
Events that together cover the entire sample space.
E1 union E2 union ... union En = S
Independent Events
Two events A and B are independent if the occurrence of one does not affect the probability of the other.
P(A and B) = P(A) x P(B) (if independent)
Example: Two separate coin tosses -- the result of the first does not affect the second.
Dependent Events
Events where the outcome of one affects the probability of the other.
Example: Drawing two cards from a deck without replacement.
5. Addition Rule (OR Rule)
For Mutually Exclusive Events
P(A or B) = P(A) + P(B)
For Non-Mutually Exclusive Events (General Rule)
P(A or B) = P(A) + P(B) - P(A and B)
For Three Events
P(A or B or C) = P(A) + P(B) + P(C) - P(A and B) - P(B and C) - P(A and C) + P(A and B and C)
Example
From a deck of 52 cards, one card is drawn. What is the probability of drawing a King or a Heart?
P(King) = 4/52
P(Heart) = 13/52
P(King and Heart) = 1/52 (King of Hearts)
P(King or Heart) = 4/52 + 13/52 - 1/52 = 16/52 = 4/13
6. Multiplication Rule (AND Rule)
For Independent Events
P(A and B) = P(A) x P(B)
For Dependent Events
P(A and B) = P(A) x P(B|A)
where P(B|A) is the probability of B given that A has occurred.
Extended to Multiple Events
P(A and B and C) = P(A) x P(B|A) x P(C|A and B)
For independent events:
P(A and B and C) = P(A) x P(B) x P(C)
Example
A bag has 5 red and 3 blue balls. Two balls are drawn without replacement. What is the probability that both are red?
P(1st red) = 5/8
P(2nd red | 1st red) = 4/7
P(both red) = 5/8 x 4/7 = 20/56 = 5/14
7. Independent vs Dependent Events
How to Identify
Independent:
- Events on separate experiments (separate coins, separate dice)
- Drawing WITH replacement
- Keyword: "and then independently"
Dependent:
- Events on the same experiment
- Drawing WITHOUT replacement
- The pool of items changes after the first draw
Formal Test
Events A and B are independent if and only if:
P(A and B) = P(A) x P(B)
If this equality does NOT hold, the events are dependent.
Example: Independent
Two dice are rolled. What is the probability of getting 6 on the first AND 4 on the second?
P(6 on 1st) = 1/6
P(4 on 2nd) = 1/6 (independent -- different dice)
P(both) = 1/6 x 1/6 = 1/36
Example: Dependent
A box has 4 white and 6 black balls. Two balls are drawn without replacement. What is the probability of getting 1 white and 1 black (in that order)?
P(1st white) = 4/10 = 2/5
P(2nd black | 1st white) = 6/9 = 2/3
P(white then black) = 2/5 x 2/3 = 4/15
8. Conditional Probability
Definition
The probability of event A occurring given that event B has already occurred:
P(A|B) = P(A and B) / P(B) [P(B) != 0]
Rearranged Forms
P(A and B) = P(B) x P(A|B) = P(A) x P(B|A)
Bayes' Theorem (Introductory)
P(A|B) = [P(B|A) x P(A)] / P(B)
Example
In a class of 100 students, 40 study Mathematics, 30 study Physics, and 10 study both. If a student is known to study Mathematics, what is the probability that they also study Physics?
P(M) = 40/100 = 2/5
P(P) = 30/100 = 3/10
P(M and P) = 10/100 = 1/10
P(P|M) = P(M and P) / P(M)
= (1/10) / (2/5)
= (1/10) x (5/2)
= 5/20
= 1/4
9. Dice Problems
Single Die
Sample space: {1, 2, 3, 4, 5, 6}
Total outcomes = 6
P(even) = 3/6 = 1/2
P(odd) = 3/6 = 1/2
P(prime) = 3/6 = 1/2 {2, 3, 5}
P(>4) = 2/6 = 1/3 {5, 6}
P(multiple of 3) = 2/6 = 1/3 {3, 6}
Two Dice
Total outcomes = 6 x 6 = 36
Sum Table for Two Dice:
Sum: 2 3 4 5 6 7 8 9 10 11 12
Ways: 1 2 3 4 5 6 5 4 3 2 1
Common Probabilities:
P(sum = 7) = 6/36 = 1/6
P(sum = 2) = 1/36
P(sum = 12) = 1/36
P(doublet) = 6/36 = 1/6 {(1,1),(2,2),...,(6,6)}
P(sum >= 10) = (3+2+1)/36 = 6/36 = 1/6
P(sum < 5) = (1+2+3)/36 = 6/36 = 1/6
Three Dice
Total outcomes = 6^3 = 216
10. Card Problems
Standard Deck Composition
Total cards = 52
4 Suits:
- Hearts (red) : 13 cards (A,2,3,...,10,J,Q,K)
- Diamonds (red) : 13 cards
- Clubs (black) : 13 cards
- Spades (black) : 13 cards
Face cards (J, Q, K): 12 total (3 per suit)
Number cards (A-10): 40 total (10 per suit)
Aces: 4
Kings: 4
Queens: 4
Jacks: 4
Red cards: 26
Black cards: 26
Single Card Drawn
P(Ace) = 4/52 = 1/13
P(King) = 4/52 = 1/13
P(Face card) = 12/52 = 3/13
P(Heart) = 13/52 = 1/4
P(Red) = 26/52 = 1/2
P(Black King) = 2/52 = 1/26
P(Red Ace) = 2/52 = 1/26
P(Queen of Spades) = 1/52
Two Cards Drawn (Without Replacement)
Total ways = 52C2 = 1,326
P(both Aces) = 4C2 / 52C2 = 6/1326 = 1/221
P(both Kings) = 4C2 / 52C2 = 1/221
P(both red) = 26C2 / 52C2 = 325/1326 = 25/102
P(one red, one black) = (26C1 x 26C1) / 52C2 = 676/1326 = 26/51
Five Cards Drawn (Poker Hands)
Total 5-card hands = 52C5 = 2,598,960
Royal Flush: 4
Straight Flush: 36
Four of a Kind: 624
Full House: 3,744
Flush: 5,108
Straight: 10,200
Three of a Kind: 54,912
Two Pair: 123,552
One Pair: 1,098,240
High Card: 1,302,540
11. Coin Problems
Single Coin
S = {H, T}
P(Head) = 1/2
P(Tail) = 1/2
n Coin Tosses
Total outcomes = 2^n
P(exactly k heads in n tosses) = nCk x (1/2)^n
= nCk / 2^n
Common Results for n Coin Tosses
n=2: S = {HH, HT, TH, TT} Total = 4
P(exactly 1 head) = 2/4 = 1/2
P(at least 1 head) = 3/4
P(both heads) = 1/4
n=3: Total = 8
P(exactly 2 heads) = 3C2/8 = 3/8
P(at least 1 head) = 1 - 1/8 = 7/8
P(all heads) = 1/8
n=4: Total = 16
P(exactly 2 heads) = 4C2/16 = 6/16 = 3/8
P(at least 1 head) = 1 - 1/16 = 15/16
n=5: Total = 32
P(exactly 3 heads) = 5C3/32 = 10/32 = 5/16
P(at least 1 head) = 1 - 1/32 = 31/32
Biased Coin
If P(Head) = p and P(Tail) = q = 1 - p, then for n tosses:
P(exactly k heads) = nCk x p^k x q^(n-k)
This is the Binomial Distribution formula.
12. Odds
Odds in Favour
Odds in favour of E = P(E) / P(E')
= Favourable outcomes : Unfavourable outcomes
Odds Against
Odds against E = P(E') / P(E)
= Unfavourable outcomes : Favourable outcomes
Converting Between Odds and Probability
If odds in favour are a : b, then:
P(E) = a / (a + b)
P(E') = b / (a + b)
Example
The odds of winning a game are 3:5. What is the probability of winning?
P(winning) = 3 / (3 + 5) = 3/8
13. Expected Value (Introduction)
Definition
If an experiment has outcomes x1, x2, ..., xn with probabilities p1, p2, ..., pn, then:
E(X) = x1*p1 + x2*p2 + ... + xn*pn
= Sum of (outcome x its probability)
Example
A game pays Rs. 10 if you roll a 6 on a die, and you lose Rs. 2 otherwise. What is the expected value?
E(X) = 10 x (1/6) + (-2) x (5/6)
= 10/6 - 10/6
= 0
The expected value is Rs. 0 (fair game).
14. Summary of All Key Formulas
+----------------------------------------------------+------------------------------------+
| Concept | Formula |
+----------------------------------------------------+------------------------------------+
| Basic Probability | P(E) = n(E) / n(S) |
| Complement | P(E') = 1 - P(E) |
| Addition (Mutually Exclusive) | P(A or B) = P(A) + P(B) |
| Addition (General) | P(AuB) = P(A)+P(B)-P(AnB) |
| Multiplication (Independent) | P(A and B) = P(A) x P(B) |
| Multiplication (Dependent) | P(A and B) = P(A) x P(B|A) |
| Conditional Probability | P(A|B) = P(AnB) / P(B) |
| Bayes' Theorem | P(A|B) = P(B|A)P(A) / P(B) |
| n coin tosses: exactly k heads | nCk / 2^n |
| Binomial (biased coin) | nCk x p^k x q^(n-k) |
| Two dice: total outcomes | 36 |
| 52-card deck: choosing r cards | 52Cr total ways |
| Odds to probability (a:b in favour) | P = a/(a+b) |
| Expected value | E(X) = Sum(xi x pi) |
+----------------------------------------------------+------------------------------------+
15. Important Identities and Properties
1. 0 <= P(E) <= 1
2. P(S) = 1, P(empty) = 0
3. P(A') = 1 - P(A)
4. P(A or B) <= P(A) + P(B)
5. P(A and B) <= min(P(A), P(B))
6. If A is a subset of B, then P(A) <= P(B)
7. P(A - B) = P(A) - P(A and B)
8. P(A' and B') = 1 - P(A or B) [De Morgan]
9. P(A' or B') = 1 - P(A and B) [De Morgan]
10. For independent events: P(A|B) = P(A)