Probability measures how likely an event is to occur. It's the foundation of predicting outcomes in uncertain situations—from coin flips to weather forecasts to medical diagnoses.
Imagine you have a bag of marbles. Probability answers: "If I pick one without looking, what are my chances of getting a red marble?"
We count:
All possible outcomes (the sample space)Outcomes we care about (favourable outcomes)Probability = fraction of favourable outcomes out of all possible ones
It's just organized counting + division.
Definition Sample Space (S)The sample space is the set of all possible outcomes of an experiment or random process.
Notation: S S S Ω \Omega Ω
WHY it matters: Before calculating probability, we need to know every possible thing that could happen. Missing outcomes breaks the calculation.
Experiment
Sample Space
Number of outcomes
Toss 1 coin
S = { H , T } S = \{\text{H}, \text{T}\} S = { H , T } 2
Roll1 die
S = { 1 , 2 , 3 , 4 , 5 , 6 } S = \{1, 2, 3, 4, 5, 6\} S = { 1 , 2 , 3 , 4 , 5 , 6 } 6
Toss 2 coins
S = { HH , HT , TH , T } S = \{\text{HH}, \text{HT}, \text{TH}, \text{T}\} S = { HH , HT , TH , T } 4
Pick a card
S = { 52 different cards } S = \{52\text{ different cards}\} S = { 52 different cards } 52
Worked example Building a Sample Space for2 DiceQuestion: What is the sample space when rolling two dice?
Step 1 — WHY: Each die has 6 outcomes. They're independent, so we need all combinations .
Step 2 — HOW: First die can show 1–6, and for each , the second die can show 1–6.
Total outcomes = 6 × 6 = 36 6 \times 6 = 36 6 × 6 = 36 Step 3 — WHAT:
S = { ( 1 , 1 ) , ( 1 , 2 ) , … , ( 1 , 6 ) , ( 2 , 1 ) , ( 2 , 2 ) , … , ( 6 , 6 ) } S = \{(1,1), (1,2), \ldots, (1,6), (2,1), (2,2), \ldots, (6,6)\} S = {( 1 , 1 ) , ( 1 , 2 ) , … , ( 1 , 6 ) , ( 2 , 1 ) , ( 2 , 2 ) , … , ( 6 , 6 )}
The sample space has36 ordered pairs.
An event is any subset of the sample space —a collection of one or more outcomes we're interested in.
Types:
Simple event: Contains exactly1 outcome (e.g., rolling a 3)Compound event: Contains multiple outcomes (e.g., rolling an even number)Sure event: E = S E = S E = S Impossible event: E = ∅ E = \emptyset E = ∅
For rolling a single die with S = { 1 , 2 , 3 , 4 , 5 , 6 } S = \{1, 2, 3, 4, 5, 6\} S = { 1 , 2 , 3 , 4 , 5 , 6 }
Event Description
Event Set
Type
"Roll a 4"
E 1 = { 4 } E_1 = \{4\} E 1 = { 4 } Simple
"Roll an even number"
E 2 = { 2 , 4 , 6 } E_2 = \{2, 4, 6\} E 2 = { 2 , 4 , 6 } Compound
"Roll less than 7"
E 3 = { 1 , 2 , 3 , 4 , 5 , 6 } = S E_3 = \{1,2,3,4,5,6\} = S E 3 = { 1 , 2 , 3 , 4 , 5 , 6 } = S Sure event
"Roll a 7"
E 4 = ∅ E_4 = \emptyset E 4 = ∅ Impossible
Start with the axioms of probability:
Axiom 1: For any event E E E 0 ≤ P ( E ) ≤ 1 0 \leq P(E) \leq 1 0 ≤ P ( E ) ≤ 1 Axiom 2: P ( S ) = 1 P(S) = 1 P ( S ) = 1 Axiom 3: If events are mutually exclusive, P ( A ∪ B ) = P ( A ) + P ( B ) P(A \cup B) = P(A) + P(B) P ( A ∪ B ) = P ( A ) + P ( B )
For equally likely outcomes:
If sample space has n n n something happening is 1:
P ( outcome 1 ) + P ( outcome 2 ) + ⋯ + P ( outcome n ) = 1 P(\text{outcome}_1) + P(\text{outcome}_2) + \cdots + P(\text{outcome}_n) = 1 P ( outcome 1 ) + P ( outcome 2 ) + ⋯ + P ( outcome n ) = 1
Since they're all equal, let P ( each outcome ) = p P(\text{each outcome}) = p P ( each outcome ) = p
n ⋅ p = 1 ⟹ p = 1 n n \cdot p = 1 \implies p = \frac{1}{n} n ⋅ p = 1 ⟹ p = n 1
Each outcome has probability 1 n ( S ) \frac{1}{n(S)} n ( S ) 1
Now if event E E E k k k
P ( E ) = 1 n ( S ) + 1 n ( S ) + ⋯ + 1 n ( S ) ⏟ k times = k n ( S ) = n ( E ) n ( S ) P(E) = \underbrace{\frac{1}{n(S)} + \frac{1}{n(S)} + \cdots + \frac{1}{n(S)}}_{k \text{ times}} = \frac{k}{n(S)} = \frac{n(E)}{n(S)} P ( E ) = k times n ( S ) 1 + n ( S ) 1 + ⋯ + n ( S ) 1 = n ( S ) k = n ( S ) n ( E )
This is why counting works for probability!
Worked example Using the ComplementQuestion: A bag has 5 red and 3 blue marbles. What's the probability of NOT picking red?
Method 1 (Direct):
P ( blue ) = 3 8 P(\text{blue}) = \frac{3}{8} P ( blue ) = 8 3 Method 2 (Complement): P ( red ) = 5 8 P(\text{red}) = \frac{5}{8} P ( red ) = 8 5 P ( not red ) = 1 − 5 8 = 3 8 P(\text{not red}) = 1 - \frac{5}{8} = \frac{3}{8} P ( not red ) = 1 − 8 5 = 8 3
WHY use complement? When counting "not E" is easier than counting E E E
Worked example Example 1: Single Die RollQuestion: What is the probability of rolling a number greater than 4?
Step 1 — Identify sample space:
S = { 1 , 2 , 3 , 4 , 5 , 6 } S = \{1, 2, 3, 4, 5, 6\} S = { 1 , 2 , 3 , 4 , 5 , 6 } n ( S ) = 6 n(S) = 6 n ( S ) = 6
WHY? The die has 6 faces, each equally likely.
Step 2 — Define the event:
E = { numbers > 4 } = { 5 , 6 } E = \{\text{numbers} > 4\} = \{5, 6\} E = { numbers > 4 } = { 5 , 6 } n ( E ) = 2 n(E) = 2 n ( E ) = 2
WHY? Only 5 and 6 satisfy "greater than 4".
Step 3 — Apply formula:
P ( E ) = n ( E ) n ( S ) = 2 6 = 1 3 P(E) = \frac{n(E)}{n(S)} = \frac{2}{6} = \frac{1}{3} P ( E ) = n ( S ) n ( E ) = 6 2 = 3 1
Answer: 1 3 \frac{1}{3} 3 1
Worked example Example 2: Two CoinsQuestion: What's the probability of getting exactly one head when tossing 2 coins?
Step 1 — Sample space:
S = { HH , HT , TH , TT } S = \{\text{HH}, \text{HT}, \text{TH}, \text{TT}\} S = { HH , HT , TH , TT } n ( S ) = 4 n(S) = 4 n ( S ) = 4 WHY these outcomes? First coin: H or T. For each, second coin: H or T. That's 2 × 2 = 4 2 \times 2 = 4 2 × 2 = 4
WHY HT and TH are different? They represent different outcomes (first coin vs. second coin showing heads).
Step 2 — Event:
E = { HT , TH } E = \{\text{HT}, \text{TH}\} E = { HT , TH } n ( E ) = 2 n(E) = 2 n ( E ) = 2 WHY? "Exactly one head" means one H and one T. That's HT or TH.
Step 3 — Calculate:
P ( E ) = 2 4 = 1 2 P(E) = \frac{2}{4} = \frac{1}{2} P ( E ) = 4 2 = 2 1 Answer: 1 2 \frac{1}{2} 2 1
Worked example Example 3: Deck of CardsQuestion: What's the probability of drawing a face card (J, Q, K) from a standard deck?
Step 1 — Sample space:
Standard deck has 52 cards, so n ( S ) = 52 n(S) = 52 n ( S ) = 52
Step 2 — Count favourable outcomes:
4 suits × 3 face cards per suit = 12 face cards
n ( E ) = 12 n(E) = 12 n ( E ) = 12 WHY 3 per suit? Jack, Queen, King in each of hearts, diamonds, clubs, spades.
Step 3 — Calculate:
P ( E ) = 12 52 = 3 13 P(E) = \frac{12}{52} = \frac{3}{13} P ( E ) = 52 12 = 13 3
Answer: 3 13 ≈ 0.231 \frac{3}{13} \approx 0.231 13 3 ≈ 0.231
Worked example Example 4: Sum of Two DiceQuestion: What's the probability that the sum of two dice equals 7?
Step 1 — Sample space:
n ( S ) = 36 n(S) = 36 n ( S ) = 36
Step 2 — Find favourable outcomes:
Sum = 7 when: ( 1 , 6 ) , ( 2 , 5 ) , ( 3 , 4 ) , ( 4 , 3 ) , ( 5 , 2 ) , ( 6 , 1 ) (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) ( 1 , 6 ) , ( 2 , 5 ) , ( 3 , 4 ) , ( 4 , 3 ) , ( 5 , 2 ) , ( 6 , 1 ) WHY these? We need pairs that add to 7. Check systematically:
If first die = 1, second must be 6
If first die = 2, second must be 5
Continue for all 6 possibilities
n ( E ) = 6 n(E) = 6 n ( E ) = 6
Step 3 — Calculate:
P ( E ) = 6 36 = 1 6 P(E) = \frac{6}{36} = \frac{1}{6} P ( E ) = 36 6 = 6 1
Answer: 1 6 ≈ 0.167 \frac{1}{6} \approx 0.167 6 1 ≈ 0.167
Common mistake Mistake 1: Forgetting Outcomes Are Not Always Equally LikelyWrong thinking: "When I toss two coins, I get 0, 1, or 2 heads—3 outcomes. So P(1head) = 1/3."
Why it feels right: We counted three distinct possibilities.
The fix: Those outcomes aren't equally likely!
0 heads: only T (1 way)
1 head: HT or TH (2 ways)
2 heads: only HH (1 way)
Correct sample space: {HH, HT, TH, TT} — these4 outcomes ARE equally likely.
Rule: Use the formula n ( E ) n ( S ) \frac{n(E)}{n(S)} n ( S ) n ( E ) S S S
Common mistake Mistake 2: Confusing "Or" with "And (Compound Events)Example: Roll a die. What's P(even OR greater than 4)?
Wrong: Count evens: {2,4,6} = 3. Count >4: {5,6} = 2. So 3 + 2 6 = 5 6 \frac{3+2}{6} = \frac{5}{6} 6 3 + 2 = 6 5
Why it's wrong: We counted 6 twice! It's even AND >4.
The fix:
Even: {2, 4, 6}
Greater than 4: {5, 6}
Even OR >4: {2, 4, 5, 6} — union without duplication
P ( E ) = 4 6 = 2 3 P(E) = \frac{4}{6} = \frac{2}{3} P ( E ) = 6 4 = 3 2 Rule: For "or", take the union of sets (no double-counting). For "and", take the intersection.
Common mistake Mistake 3: Wrong Sample Space for Ordered vs. UnorderedScenario: Pick 2 balls from {R, B, G} without replacement.
Wrong sample space: {RB, RG, BG} — 3 outcomes, so P(getting R) = 2/3?
Why it feels right: We're just looking at pairs.
The fix: Are we distinguishing order?
Ordered (which came first matters): {RB, BR, RG, GR, BG, GB} — 6 outcomesUnordered (just the pair matters): {RB, RG, BG} — 3 outcomes, BUT probabilities differ!
For equally likely picking: use ordered. Each pick sequence is equally likely.
Rule: When objects are drawn sequentially, distinguish order unless explicitly told outcomes are "combinations."
Recall Feynman Technique — Explain to a 12-Year-OldImagine you're explaining probability to your younger sibling:
"Okay, so probability is like guessing games. Say you have a box with 10 toys—7 cars and 3 dolls. You close your eyes and grab one. What are your chances of getting a car?
First, count ALL toys: 10 total. That's everything that could happen.
Next, count what you WANT: 7 cars.
Your chances = the fraction: 7 out of 10, or 7/10. That's 0.7, or 70%.
If I said 'What's the chance of NOT getting a car?' — well, that's getting a doll. 3 dols out of 10 toys = 3/10 or 30%.
Notice: 70% + 30% = 100%. Because you MUST get something!
Probability is just: (what you want) ÷ (all possibilities). As long as every toy has an equal chance of being picked, this works perfectly."
Mnemonic Remember the Formula"FavoURable over TOTal"
P = F T P = \frac{\text{F}}{\text{T}} P = T F F avourable / T otal
Or think: "PART over WHOLE" — the part you want over the whole set of options.
Sample Space — all possible outcomesEvents and Set Theory — events as subsets, unions, intersectionsComplement of an Event — what doesn't happenMutually Exclusive Events — can't happen togetherAxioms of Probability — the foundation rulesConditional Probability — probability given extra informationCounting Principles — permutations and combinations for complex sample spacesRandom Variables — numerical outcomes and their probabilitiesLaw of Large Numbers — probability → frequency with many trials
#flashcards/maths
What is a sample space? :: The set of all possible outcomes of a random experiment.
What is an event in probability? A subset of the sample space; a collection of one or more outcomes we're interested in.
What is the classical probability formula? P ( E ) = n ( E ) n ( S ) P(E) = \frac{n(E)}{n(S)} P ( E ) = n ( S ) n ( E ) where
n ( E ) n(E) n ( E ) is the number of favourable outcomes and
n ( S ) n(S) n ( S ) is the total number of outcomes, assuming all outcomes are equally likely.
What is the range of any probability value? 0 ≤ P ( E ) ≤ 1 0 \leq P(E) \leq 1 0 ≤ P ( E ) ≤ 1 . A probability cannot be negative or greater than 1.
What is the complement rule? P ( E ′ ) = 1 − P ( E ) P(E') = 1 - P(E) P ( E ′ ) = 1 − P ( E ) where
E ′ E' E ′ is the event that
E E E does not occur.
If you roll a fair die, what is P(rolling a number ≤ 2)? 2 6 = 1 3 \frac{2}{6} = \frac{1}{3} 6 2 = 3 1 because favourable outcomes are {1, 2} out of 6 total outcomes.
When tossing two coins, what is the sample space? {HH, HT, TH, TT} — four equally likely outcomes.
What is P(exactly one head) when tossing two fair coins? 2 4 = 1 2 \frac{2}{4} = \frac{1}{2} 4 2 = 2 1 because {HT, TH} are the favourable outcomes out of 4 total.
How many outcomes are in the sample space for rolling two dice? 36 outcomes (6 × 6), represented as ordered pairs (1,1) through (6,6).
What is P(sum = 7) when rolling two dice? 6 36 = 1 6 \frac{6}{36} = \frac{1}{6} 36 6 = 6 1 because there are 6 ways to get sum 7: {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}.
Why can't we always just count outcomes and divide? The formula
P ( E ) = n ( E ) n ( S ) P(E) = \frac{n(E)}{n(S)} P ( E ) = n ( S ) n ( E ) only works when all outcomes in the sample space are equally likely.
In a deck of 52 cards, what is P(drawing a heart)? 13 52 = 1 4 \frac{13}{52} = \frac{1}{4} 52 13 = 4 1 because there are 13 hearts in52 cards.
Favourable outcomes n of E
Intuition Hinglish mein samjhoDekho, probability matlab ye samajhne ki koshish hai kioi chez hone ka kitna chance hai. Jaise agar ek bag mein 10 marbles hain — 7 laal aur 3 nele — aur tum bina dekhe ek uthate ho, toh laal marble milne ka chance kya hai? Simple: 7 laal hain total 10 mein se, toh probability = 7/10 ya 70%.
Pehle hum sample space banate hain — matlab sab possible outcomes ko list karte hain. Ek die roll karo toh sample space hai {1, 2, 3, 4, 5, 6}. Do coins toss karo toh {HH, HT, TH, TT} — yahan 4 outcomes hain, kyunki pehla coin aur dosra coin alag-alag outcomes de sakte hain. Phir hum apna event define karte hain — matlab hume kya chahiye. Agar event hai "4 se bada number", toh outcomes {5, 6} hain. Uske baad formula lagao: P(E) = (favourable outcomes) / (total outcomes) = 2/6 = 1/3.
Ek important baat: ye formula tab hi kaam karta hai jab har outcome equally likely ho . Agar do coins toss karke tum socho ki 0head, 1 head, 2 head — teen outcomes hain toh P(1 head) = 1/3, toh galat hai! Actual sample space {HH, HT, TH, TT} hai jisme 1 head ke liye HT aur TH dono favourable hain, toh 2/4 = 1/2. Probability mein accurate counting bohot zaroori hai — warna answer galat aa jayega. Yahi basic foundation hai pore probability ka!
