Episode 8 — Aptitude and Reasoning / 8.15 — Probability
8.15 Quick Revision -- Probability
Core Formulas at a Glance
+----------------------------------------------------+------------------------------------+
| Concept | Formula |
+----------------------------------------------------+------------------------------------+
| Basic Probability | P(E) = n(E) / n(S) |
| Range | 0 <= P(E) <= 1 |
| Certain event | P(S) = 1 |
| Impossible event | P(empty) = 0 |
| Complement | P(E') = 1 - P(E) |
+----------------------------------------------------+------------------------------------+
Addition Rule (OR)
Mutually Exclusive: P(A or B) = P(A) + P(B)
General: P(A or B) = P(A) + P(B) - P(A and B)
Three events: P(AuBuC) = P(A)+P(B)+P(C)-P(AnB)-P(BnC)-P(AnC)+P(AnBnC)
Multiplication Rule (AND)
Independent: P(A and B) = P(A) x P(B)
Dependent: P(A and B) = P(A) x P(B|A)
Three independent: P(A and B and C) = P(A) x P(B) x P(C)
Conditional Probability
P(A|B) = P(A and B) / P(B)
Bayes: P(A|B) = P(B|A) x P(A) / P(B)
Sample Spaces to Memorize
+-------------------------------+-----------------+
| Experiment | Total Outcomes |
+-------------------------------+-----------------+
| 1 coin | 2 |
| 2 coins | 4 |
| 3 coins | 8 |
| n coins | 2^n |
| 1 die | 6 |
| 2 dice | 36 |
| 3 dice | 216 |
| 1 card from 52 | 52 |
| 2 cards from 52 | 1,326 |
| 5 cards from 52 | 2,598,960 |
+-------------------------------+-----------------+
Two-Dice Sum Table
Sum: 2 3 4 5 6 7 8 9 10 11 12
Ways: 1 2 3 4 5 6 5 4 3 2 1
Prob: -- -- -- -- -- 1/6 -- -- -- -- --
1/ 1/ 1/ 1/ 5/ 5/ 1/ 1/ 1/ 1/
36 18 12 9 36 36 9 12 18 36
52-Card Deck Reference
Total = 52
Suits = 4 (Hearts, Diamonds = red; Clubs, Spades = black)
Cards per suit = 13 (A,2,3,4,5,6,7,8,9,10,J,Q,K)
Red cards = 26 Black cards = 26
Face cards = 12 Aces = 4
Kings = 4 Queens = 4
Jacks = 4 Number cards (2-10) = 36
Coin Toss Quick Reference
P(exactly k heads in n tosses of fair coin) = nCk / 2^n
n=2: P(0H)=1/4 P(1H)=1/2 P(2H)=1/4
n=3: P(0H)=1/8 P(1H)=3/8 P(2H)=3/8 P(3H)=1/8
n=4: P(0H)=1/16 P(1H)=1/4 P(2H)=3/8 P(3H)=1/4 P(4H)=1/16
P(at least 1 head in n tosses) = 1 - (1/2)^n
Binomial Probability (Biased Coin / Repeated Trials)
P(exactly k successes in n trials) = nCk x p^k x q^(n-k)
where p = probability of success, q = 1-p
Key Shortcuts
1. "At least 1" --> 1 - P(none)
2. "At least one 6 in 2 dice" --> 1 - (5/6)^2 = 11/36
3. With replacement --> events are INDEPENDENT (multiply same probabilities)
4. Without replacement --> events are DEPENDENT (probabilities change)
5. Simultaneous draws --> use combinations: P = (fav nCr) / (total nCr)
Common Probabilities to Memorize
Single die:
P(any specific number) = 1/6
P(even) = P(odd) = 1/2
P(prime: 2,3,5) = 1/2
Single card:
P(any specific card) = 1/52
P(any rank: King, Ace, etc.) = 1/13
P(any suit) = 1/4
P(face card) = 3/13
P(red) = 1/2
Two dice:
P(sum=7) = 1/6 (most likely sum)
P(doublet) = 1/6
P(sum=2) = P(sum=12) = 1/36
Coins:
P(H) = P(T) = 1/2
P(at least 1H in 3 tosses) = 7/8
P(at least 1H in 4 tosses) = 15/16
Decision Flowchart
Problem type?
|
+-- "A AND B both happen"
| +-- Independent? --> P(A) x P(B)
| +-- Dependent? --> P(A) x P(B|A)
|
+-- "A OR B (either)"
| +-- Mutually exclusive? --> P(A) + P(B)
| +-- Not exclusive? --> P(A) + P(B) - P(A and B)
|
+-- "At least one / none"
| --> Use COMPLEMENT: 1 - P(opposite)
|
+-- "Given that B happened, find P(A)"
| --> Conditional: P(A and B) / P(B)
|
+-- "Exactly k successes in n trials"
--> Binomial: nCk x p^k x q^(n-k)
Odds Conversion
Odds in favour a:b --> P(E) = a/(a+b)
Odds against a:b --> P(E) = b/(a+b)
P(E) to odds in favour: P(E) : P(E') = P(E) : (1-P(E))
Top 5 Exam Traps
1. Using WITH-replacement formula for WITHOUT-replacement problems (or vice versa).
2. Forgetting to subtract overlap in P(A or B) when events are not mutually exclusive.
3. Computing P(at least 1) directly instead of using 1 - P(none).
4. Treating (1,2) and (2,1) as the same outcome in two-dice problems.
5. Not updating probabilities after each draw in sequential without-replacement problems.
De Morgan's Laws for Probability
P(A' and B') = P(neither A nor B) = 1 - P(A or B)
P(A' or B') = P(not both) = 1 - P(A and B)
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