Episode 8 — Aptitude and Reasoning / 8.15 — Probability
8.15.b Tips, Tricks, and Shortcuts -- Probability
1. The First Step in Every Problem
Before anything else, answer these three questions:
1. What is the EXPERIMENT? (tossing coins, drawing cards, rolling dice, etc.)
2. What is the SAMPLE SPACE? (list or count all possible outcomes)
3. What is the EVENT? (which outcomes count as "favourable"?)
Getting the sample space wrong is the number one source of errors.
2. When to Use Which Rule
Decision Guide
Problem says "A AND B" (both happen)?
|
+-- Are A and B independent?
| YES --> P(A and B) = P(A) x P(B)
| NO --> P(A and B) = P(A) x P(B|A)
|
Problem says "A OR B" (either happens)?
|
+-- Are A and B mutually exclusive?
YES --> P(A or B) = P(A) + P(B)
NO --> P(A or B) = P(A) + P(B) - P(A and B)
Keyword Mapping
AND keywords: both, all, simultaneously, together, each
--> MULTIPLY probabilities
OR keywords: either, at least one, any one, one of
--> ADD probabilities (with correction for overlap)
COMPLEMENT keywords: at least 1, none, not, neither
--> Use P(E) = 1 - P(E')
3. The Complement Trick (Most Powerful Shortcut)
When to Use It
Whenever a problem says "at least one", "at least 1", or the direct counting looks complex, use:
P(at least 1) = 1 - P(none)
Why It Works
"At least one" includes many cases (exactly 1, exactly 2, ..., all). The complement "none" is usually just one simple case.
Example
Probability that at least one of three independently shot arrows hits the target, if each has probability 1/3 of hitting:
Direct approach: P(exactly 1) + P(exactly 2) + P(exactly 3) -- messy!
Complement approach:
P(none hits) = (2/3)^3 = 8/27
P(at least 1 hits) = 1 - 8/27 = 19/27
4. With vs Without Replacement
This distinction is CRITICAL and often missed.
WITH REPLACEMENT:
- The pool resets after each draw
- Events are INDEPENDENT
- Probabilities stay the same for each draw
- Total outcomes: n^r (n items, r draws)
WITHOUT REPLACEMENT:
- The pool shrinks after each draw
- Events are DEPENDENT
- Probabilities change with each draw
- Total outcomes: nPr or nCr depending on order
Quick Example
Bag: 3 red, 7 blue. Draw 2 balls. P(both red)?
With replacement: 3/10 x 3/10 = 9/100
Without replacement: 3/10 x 2/9 = 6/90 = 1/15
5. Two-Dice Shortcuts
Memorize the Sum Distribution
Sum | Ways | Probability
-----|------|------------
2 | 1 | 1/36
3 | 2 | 2/36 = 1/18
4 | 3 | 3/36 = 1/12
5 | 4 | 4/36 = 1/9
6 | 5 | 5/36
7 | 6 | 6/36 = 1/6 <-- most likely sum
8 | 5 | 5/36
9 | 4 | 4/36 = 1/9
10 | 3 | 3/36 = 1/12
11 | 2 | 2/36 = 1/18
12 | 1 | 1/36
Key Pattern
Number of ways to get sum k:
- For k = 2 to 7: ways = k - 1
- For k = 8 to 12: ways = 13 - k
Quick Calculations
P(sum is even) = 18/36 = 1/2
P(sum is odd) = 18/36 = 1/2
P(doublet) = 6/36 = 1/6
P(sum > 9) = (3+2+1)/36 = 6/36 = 1/6
P(sum = 7 or 11) = (6+2)/36 = 8/36 = 2/9
P(at least one 6) = 1 - P(no 6) = 1 - (5/6)^2 = 11/36
6. Card Problem Shortcuts
Know the Deck Cold
Total: 52
Each suit: 13 (A,2,3,4,5,6,7,8,9,10,J,Q,K)
Red suits: Hearts, Diamonds (26 cards)
Black suits: Clubs, Spades (26 cards)
Face cards: J,Q,K (12 total, 3 per suit)
Aces: 4
Numbered cards (2-10): 36 total
Common Single-Card Probabilities
P(specific card) = 1/52
P(specific rank) = 4/52 = 1/13
P(specific suit) = 13/52 = 1/4
P(face card) = 12/52 = 3/13
P(red face card) = 6/52 = 3/26
P(not a face card) = 40/52 = 10/13
Combination-Based Card Problems
When multiple cards are drawn simultaneously, use:
P = Favourable combinations / Total combinations
= (ways to choose desired cards) / 52Cr
7. Coin Problem Shortcuts
Formula for n Fair Coins
P(exactly k heads) = nCk / 2^n
Pre-Computed Values for Quick Reference
n=1: P(1H) = 1/2
n=2: P(0H) = 1/4 P(1H) = 2/4 = 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) = 4/16 P(2H) = 6/16 P(3H) = 4/16 P(4H) = 1/16
n=5: P(0H) = 1/32 P(1H) = 5/32 P(2H) = 10/32 P(3H) = 10/32 P(4H) = 5/32 P(5H) = 1/32
"At Least" Shortcut
P(at least k heads in n tosses) = Sum of nCi/2^n for i = k to n
Or equivalently:
P(at least 1 head) = 1 - (1/2)^n = (2^n - 1) / 2^n
8. Tree Diagram Strategy
For multi-stage problems, draw a tree:
Example: Bag has 3R, 2B. Draw 2 without replacement.
Start
/ \
3/5 2/5
/ \
R1 B1
/ \ / \
2/4 2/4 3/4 1/4
/ \ / \
R2 B2 R2 B2
P(RR) = 3/5 x 2/4 = 6/20 = 3/10
P(RB) = 3/5 x 2/4 = 6/20 = 3/10
P(BR) = 2/5 x 3/4 = 6/20 = 3/10
P(BB) = 2/5 x 1/4 = 2/20 = 1/10
Check: 3/10 + 3/10 + 3/10 + 1/10 = 10/10 = 1 (correct)
When to Use Trees
- Problems with 2-3 sequential stages
- When different branches have different probabilities
- When the problem involves dependent events
9. "Or" Problems: When to Add vs When Not To
Mutually Exclusive -- Just Add
P(rolling 2 or 5 on a die) = 1/6 + 1/6 = 2/6 = 1/3
(You cannot roll both 2 and 5 on a single roll)
Not Mutually Exclusive -- Subtract the Overlap
P(Heart or King from a deck):
P(Heart) = 13/52, P(King) = 4/52, P(Heart AND King) = 1/52
P(Heart or King) = 13/52 + 4/52 - 1/52 = 16/52 = 4/13
Quick Test
Can both events happen at the same time?
- No --> Mutually Exclusive --> Just add
- Yes --> Not Mutually Exclusive --> Add and subtract overlap
10. Geometric Probability (Bonus)
When the sample space is continuous (e.g., a region), use:
P(E) = Favourable area (or length) / Total area (or length)
Example: A point is chosen at random inside a square of side 10. What is the probability it lies inside an inscribed circle?
Area of circle = pi x 5^2 = 25*pi
Area of square = 10^2 = 100
P = 25*pi / 100 = pi/4 ≈ 0.785
11. Common Traps and How to Avoid Them
Trap 1: Treating Dependent Events as Independent
WRONG: P(2 aces from 52 cards without replacement) = (4/52) x (4/52)
RIGHT: P(2 aces without replacement) = (4/52) x (3/51) = 12/2652 = 1/221
Or: 4C2 / 52C2 = 6/1326 = 1/221
Trap 2: Forgetting to Use Complement
HARD WAY: P(at least 2 heads in 5 tosses) = P(2H) + P(3H) + P(4H) + P(5H)
EASY WAY: 1 - P(0H) - P(1H)
= 1 - 1/32 - 5/32
= 1 - 6/32
= 26/32
= 13/16
Trap 3: Confusing "OR" with "AND"
"Both dice show 6" --> AND --> multiply
"At least one die shows 6" --> OR (complement) --> 1 - P(neither shows 6)
Trap 4: Order Matters in Sequential Drawing
P(first red then blue) != P(one red and one blue)
P(one red and one blue) = P(RB) + P(BR) <-- both orders
Trap 5: Miscounting the Sample Space
Two dice: outcomes = 36 (not 12, not 21)
(1,2) and (2,1) are different outcomes!
Two coins: outcomes = 4 (HH, HT, TH, TT)
HT and TH are different!
12. Speed Techniques for Exams
Technique 1: Convert to Fractions Early
Work entirely in fractions. Avoid decimals until the final answer.
Technique 2: Cross-Cancel Before Multiplying
P = (4/52) x (3/51)
= (1/13) x (1/17) [cancel 4 with 52, cancel 3 with 51]
= 1/221
Technique 3: Use nCr for Simultaneous Draws
When cards/balls are drawn "simultaneously" or "at once," use combinations directly:
P = (favourable nCr) / (total nCr)
This avoids order considerations entirely.
Technique 4: Symmetry Arguments
If the problem is symmetric, use it:
P(first card is Ace) = P(any specific card is Ace) = 4/52 = 1/13
(True regardless of how many cards are drawn, if we only care about one position)
Technique 5: Sanity Checks
- Probability must be between 0 and 1
- All probabilities in a sample space must sum to 1
- P(at least 1) >= P(exactly 1)
- P(A or B) >= max(P(A), P(B))
- P(A and B) <= min(P(A), P(B))
13. Formula Quick-Select Guide
Problem Type | Formula to Use
------------------------------------|----------------------------------
Single event, simple counting | P = favourable / total
At least one / none | 1 - P(complement)
Both A and B, independent | P(A) x P(B)
Both A and B, dependent | P(A) x P(B|A)
Either A or B, exclusive | P(A) + P(B)
Either A or B, not exclusive | P(A) + P(B) - P(A and B)
Exactly k successes in n trials | nCk x p^k x q^(n-k)
Conditional probability | P(A and B) / P(B)
Drawing r from n items | nCr-based counting
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