NCERT Plus Two (Class 12) Mathematics • Chapter 13

Probability

Conditional probability P(E|F), multiplication theorem, independent events, total probability & Bayes' theorem.

📄 Official NCERT Textbook PDF

Topic 13.2: Conditional Probability P(E|F) & Reduced Sample Space

Suppose you roll a fair die behind a screen, and your friend peeks and whispers a secret clue: "The number is EVEN!" Instantly {1, 3, 5} vanish from possibility — your sample space shrinks from 6 faces down to just F = {2, 4, 6}! Now what's the chance it's ALSO greater than 3 ({4, 6})? 2 out of 3! That clue-shrinking superpower is Conditional Probability P(E|F)!
CORE FORMULAP(E|F) = P(E ∩ F)P(F)  (P(F) > 0)
SHRUNKEN UNIVERSEGiven event F becomes the NEW sample space (denominator); only E ∩ F survives on top!
KEY PROPERTIESP(F|F) = 1  |  P(E'|F) = 1 − P(E|F)
P((A∪B)|F) = P(A|F) + P(B|F) − P((A∩B)|F)
🧠 Memory Trick: Whatever sits AFTER the slash | ("given that...") ALWAYS drops straight into the DENOMINATOR!
💡 Simple Kid-Check: Read P(E | F) aloud as "Chance of E GIVEN THAT F already happened!" Once F happens, ignore everything outside F — zoom your camera right into circle F and ask: "What fraction of F belongs to E?"
  • 1
    Identify E & F: Put the event you want before | (E), and the clue already known after | (F).
  • 2
    Equally Likely Shortcut: When sample points are equally likely, count directly: P(E|F) = n(E ∩ F)n(F)!
  • 3
    From Given Probabilities: Use P(E|F) = P(E ∩ F) / P(F) and P(A ∪ B) = P(A) + P(B) − P(A ∩ B) when needed.
🧒 Two Famous NCERT Classics:
• Two-Children Puzzle (Ex 2): Sample space S = {BB, BG, GB, GG} (elder first, younger second).
  (i) P(both boys | youngest is a boy): F = {BB, GB} (2 cases), E ∩ F = {BB} → ½.
  (ii) P(both boys | at least one is a boy): F = {BB, BG, GB} (3 cases!), E ∩ F = {BB} → ⅓!
• Sum = 6 on Two Dice (Ex 6): Given sum is 6 → F = {(1,5),(2,4),(3,3),(4,2),(5,1)} (5 cases). Probability 4 appeared at least once: {(2,4),(4,2)} → 2/5!
Watch TWO super-common traps! (1) Swapping P(A|B) and P(B|A) — they are NOT equal (P(A|B) divides by P(B), while P(B|A) divides by P(A))! (2) In the two-children puzzle, saying "one child is a boy, so the other is 50-50 boy/girl" — WRONG! "At least one boy" leaves 3 equally likely pairs {BB, BG, GB}, so the chance both are boys is 1/3, NOT 1/2!

Topic 13.3: Multiplication Theorem on Probability for Dependent Events

Imagine picking TWO sweets one-by-one from a jar holding 10 dark halwas and 5 white halwas in Kozhikode — WITHOUT putting the first sweet back! After you eat one dark halwa, only 14 sweets remain inside (9 dark left)! Because the second draw depends on the first draw, just rearrange the conditional formula into the Multiplication Theorem!
2-EVENT MULTIPLICATIONP(E ∩ F) = P(E) · P(F | E)
= P(F) · P(E | F)
3-EVENT CHAIN RULEP(E ∩ F ∩ G) = P(E) · P(F|E) · P(G|E∩F)
(Multiply step-by-step along the tree!)
WITHOUT REPLACEMENTBoth numerator (if same item) AND denominator shrink by 1 on each draw: 52 → 51 → 50!
🧠 Memory Trick: Multiply along the branches from left to right — ALWAYS update what's left in the bag if "without replacement"!
💡 Simple Kid-Check: Start with P(F | E) = P(E ∩ F) / P(E) and cross-multiply P(E) across — boom, P(E ∩ F) = P(E) · P(F | E)! Read it like a story: "First event E happens, AND THEN given E already happened, event F happens!"
  • 1
    Step 1 (1st Draw): Write initial probability P(E) using total items in the urn/deck.
  • 2
    Step 2 (2nd Draw): Check whether the item was replaced! If NOT replaced, reduce total count by 1 (and matching count by 1 if needed) to write P(F | E).
  • 3
    Step 3 (Multiply): Multiply all step probabilities together for P(E ∩ F) (or P(E ∩ F ∩ G)).
🧒 NCERT Examples 8 & 9 (Classic Sequential Draws):
• Two Black Balls (Ex 8): Urn has 10 Black + 5 White (15 total). Two drawn without replacement.
  P(1st Black) = 10/15. Now 9 Black & 14 total left → P(2nd Black | 1st Black) = 9/14.
  P(both Black) = 1015 × 914 = 23 × 914 = 37!
• Three Cards (Ex 9): Drawn successively without replacement; first 2 Kings, 3rd Ace:
  P = 452 × 351 × 450 = 113 × 117 × 225 = 25525!
Watch the word "without replacement" or "successively"! Students casually write (4/52) × (3/52) instead of reducing the deck to 51! Also watch "at random together" vs "one by one without replacement" — for unordered events (like "one black, one white"), remember there are TWO orders: BW OR WB!

Topic 13.4: Independent Events P(E ∩ F) = P(E)P(F) vs Mutually Exclusive

Toss a coin in Kochi and roll a die in Kozhikode! Does landing Head on the coin magically force the die to roll a 6? ZERO connection! Because knowing F happened doesn't change the odds of E (P(E|F) = P(E)), the two events are Independent — so the Multiplication Rule simplifies cleanly to P(E ∩ F) = P(E) · P(F)!
INDEPENDENCE TESTE, F are independent iff P(E ∩ F) = P(E) · P(F)! (For 3 events: check ALL 3 pairs AND P(A∩B∩C)=P(A)P(B)P(C)!)
COMPLEMENT GIFTIf E, F are independent, then (E', F), (E, F'), and (E', F') are ALSO independent!
AT-LEAST-ONE SHORTCUTP(A ∪ B) = 1 − P(A')P(B')
(Fastest way to solve "problem is solved"!)
🧠 Memory Trick: Mutually Exclusive = CANNOT happen together (P(E∩F) = 0); Independent = DON'T CARE about each other (P(E∩F) = P(E)·P(F))!
💡 Simple Kid-Check: Kids often mix up "Independent" and "Mutually Exclusive" — they are actually almost OPPOSITES! If two non-zero events are Mutually Exclusive (like rolling Odd and Even), knowing Odd happened guarantees Even did NOT happen (P(Even|Odd) = 0 ≠ 1/2) — total dependence! Independent events CAN happen together, just without influencing each other!
FeatureMutually Exclusive EventsIndependent Events
Core MeaningDisjoint sets (cannot co-occur)No influence on each other's odds
FormulaP(E ∩ F) = 0P(E ∩ F) = P(E) · P(F)
P(E ∪ F)P(E) + P(F)1 − P(E')P(F') (or P(E)+P(F)−P(E)P(F))
Can both hold?ONLY if at least one event has probability 0! (If P(E)>0, P(F)>0, mutually exclusive ⇒ NEVER independent!)
🧒 Two High-Yield Board Examples:
• Problem Solved (Ex 14): A & B try a problem independently with odds ½ and ⅓.
  (i) P(problem is solved) = 1 − P(A')P(B') = 1 − (1 − ½)(1 − ⅓) = 1 − (½)(⅔) = 1 − ⅓ = ⅔!
  (ii) P(exactly one solves) = P(A)P(B') + P(A')P(B) = (½)(⅔) + (½)(⅓) = ⅓ + ⅙ = ½!
• Find p Trap (Ex 13.2 Q7): P(A) = ½, P(A ∪ B) = 3/5, P(B) = p.
  If Mutually Exclusive: 3/5 = ½ + p − 0 ⇒ p = 1/10.
  If Independent: 3/5 = ½ + p − ½p ⇒ 1/10 = ½p ⇒ p = 1/5!
THE #1 conceptual trap in Chapter 13! Never confuse Mutually Exclusive (P(A ∩ B) = 0) with Independent (P(A ∩ B) = P(A)P(B))! Also remember for THREE events A, B, C: checking ONLY P(A∩B∩C) = P(A)P(B)P(C) is NOT enough — they must ALSO satisfy all three pairwise checks P(A∩B)=P(A)P(B), P(B∩C)=P(B)P(C), and P(A∩C)=P(A)P(C)!

Topic 13.5.1 – 13.5.2: Partition of Sample Space & Theorem of Total Probability

Imagine every student at a Kochi college arrives on exactly one of three buses — Bus 1 (E₁), Bus 2 (E₂), or Bus 3 (E₃) — nobody rides two buses and nobody walks (a Partition of S!). Some riders carry umbrellas (A). What is the overall chance P(A) a random student carries an umbrella? Just add the umbrella flow coming off ALL three buses!
PARTITION OF SE₁, ..., Eₙ are pairwise disjoint (Eᵢ∩Eⱼ=∅), exhaustive (⋃ Eᵢ = S), with P(Eᵢ) > 0.
TOTAL PROBABILITYP(A) = P(E₁)P(A|E₁) + P(E₂)P(A|E₂) + ... + P(Eₙ)P(A|Eₙ)
SANITY CHECKCheck that all branch priors add to 1: Σ P(Eᵢ) = 1!
🧠 Memory Trick: Total Probability walks FORWARD down EVERY branch of the probability tree and ADDS every branch that ends at A!
💡 Simple Kid-Check: Picture a big rectangular cake S sliced cleanly into non-overlapping strips E₁, E₂, E₃. Now pour chocolate icing A across the cake! The total chocolate area P(A) is simply the chocolate on strip 1 PLUS strip 2 PLUS strip 3: P(E₁)P(A|E₁) + P(E₂)P(A|E₂) + P(E₃)P(A|E₃)!
  • 1
    Identify the Partition E₁, E₂, ...: Pick mutually exclusive, exhaustive Stage-1 cases (bags, machines, routes) whose priors sum to 1.
  • 2
    Find Conditional Odds P(A | Eᵢ): For each strip Eᵢ, write the probability of target event A inside that strip.
  • 3
    Multiply & Sum All Paths: P(A) = Σ P(Eᵢ) · P(A | Eᵢ).
🧒 NCERT Example 15 (Construction Strike):
Probability of a strike is P(E₁) = 0.65 → no strike P(E₂) = 0.35. Probability job finishes on time given NO strike is P(A|E₂) = 0.80, and given a strike is P(A|E₁) = 0.32.
Total Probability job finishes on time:
P(A) = P(E₁)P(A|E₁) + P(E₂)P(A|E₂) = (0.65)(0.32) + (0.35)(0.80) = 0.208 + 0.280 = 0.488!
NEVER average conditional probabilities directly like (0.32 + 0.80)/2! That only works when priors happen to be equal (½, ½)! Whenever priors are unequal (like 0.65 and 0.35, or picking Bag I on die roll {1,2} → 2/6 vs Bag II on {3,4,5,6} → 4/6), ALWAYS multiply each branch by its exact weight P(Eᵢ)!

Topic 13.5: Bayes' Theorem: Reversing Conditional Causes Backward

Total Probability predicts the future: "Pick a random bag → what's the chance of drawing a Red ball?" Bayes' Theorem plays the detective traveling BACKWARD in time: you ALREADY see a Red ball (A) lying on the table! Now ask: "What is the chance this Red ball came from Bag II (E₂)?" Just divide Bag II's Red branch by the sum of ALL Red branches!
BAYES' THEOREMP(Eᵢ | A) = P(Eᵢ) · P(A | Eᵢ)Σⱼ P(Eⱼ) · P(A | Eⱼ)
PLAIN ENGLISH RULEP(Cause | Clue) = That Cause's BranchTotal Probability P(Clue)
PRIOR → POSTERIORP(Eᵢ) = Prior probability (before clue);
P(Eᵢ | A) = Posterior probability (after clue!).
🧠 Memory Trick: Target branch on TOP, ALL branches ending in the observed event summed on BOTTOM!
💡 Simple Kid-Check: Every Bayes question gives you the ending first ("a Red ball was drawn", "a Gold coin was picked", "a defective bolt was found") and asks for the starting cause ("probability it came from Bag II / Box I / Machine B"). Put the requested starting cause on TOP and Total Probability of the observed ending on BOTTOM!
  • 1
    Define Hypotheses E₁, E₂, ...: List the disjoint Stage-1 causes and write their priors P(Eᵢ).
  • 2
    Define Observed Event A: Write the conditional likelihood P(A | Eᵢ) inside each cause.
  • 3
    Plug into Bayes' Formula: Target branch P(Eₖ)P(A|Eₖ) on top ÷ sum of all branches Σ P(Eⱼ)P(A|Eⱼ) below!
🧒 NCERT Examples 16 & 17 (Must-Know Classics!):
• Two Bags (Ex 16): Bag I (3 Red, 4 Black); Bag II (5 Red, 6 Black). Bag picked at random (½ each). A Red ball is drawn → find P(from Bag II):
  P(E₁) = P(E₂) = ½,  P(A|E₁) = 3/7,  P(A|E₂) = 5/11.
  P(E₂ | A) = (½)(5/11)(½)(3/7) + (½)(5/11) = 5/11(33 + 35)/77 = 3568!
• Three Coin Boxes (Ex 17): Box I {G, G}, Box II {S, S}, Box III {G, S}. Pick a box (⅓ each), draw 1 coin → it is Gold! Probability the OTHER coin in that box is also Gold (= P(Box I | Gold)):
  P(A|E₁) = 1, P(A|E₂) = 0, P(A|E₃) = ½ ⇒ P(E₁|A) = 11 + 0 + ½ = 23! (NOT ½!)
THE #1 DHSE Bayes mistake: confusing P(A | E₂) (forward likelihood = 5/11) with P(E₂ | A) (backward Bayes posterior = 35/68)! Always ask yourself: "Which event ALREADY OCCURRED?" That observed clue ALWAYS sits AFTER the vertical slash | in P(Cause | Observed Clue)!

Topic 13.5: Real-World Bayes: Medical Testing, Factory Bolts & Truth-Teller

Suppose a medical test for a rare illness (0.1% of people have it) is 90% accurate on sick people and makes only 1% false alarms on healthy people! If you test POSITIVE, are you 90% likely to be sick? Shockingly, NO — it's only ~8.3%! Why? Because 99.9% of people are healthy, so even 1% false alarms outnumber the rare true positives! That counter-intuitive superpower makes Bayes essential in medicine, factories, and 5-mark board exams!
FACTORY DEFECT MODELMachines E₁, E₂, E₃ make 25%, 35%, 40% bolts with defect rates 5%, 4%, 2% → find P(E₂ | Defective)!
MEDICAL FALSE-POSITIVEP(Disease | +) = P(D)·P(+|D)P(D)·P(+|D) + P(D')·P(+|D')
Watch false positive P(+|D') carefully!
TRUTH-TELLER DIETruth 3/4, man reports 6: E₁ = actually 6 (1/6), E₂ = not 6 (5/6). P(E₁|reports 6) = (1/6)(3/4)(1/6)(3/4)+(5/6)(1/4) = 3/8!
🧠 Memory Trick: Convert every percentage to clean decimals (0.25 × 0.05) or hundredths (25×5 out of 10,000) — cancel denominators cleanly before dividing!
💡 Simple Kid-Check (10,000-Person Trick): Why is a 99%-specific medical test only 8.3% positive predictive in Ex 18? Imagine 10,000 people! Only 10 actually have the disease (0.1%), so 9 test positive. Meanwhile 9,990 are healthy — and 1% false positives = 100 healthy people testing positive! So out of 9 + 100 = 109 positive tests, only 9/109 ≈ 8.3% actually have the disease!
Board BlueprintPartition E₁, E₂, ...Observed Clue AKey Exam Check
Factory Bolts (Ex 19)Machines A, B, C outputsPicked bolt is defectiveΣ P(Machine) = 100% = 1
Medical Test (Ex 18)Has disease E vs Healthy E'Test reads Positive (+)Keep P(+|E') distinct from 1−P(+|E)!
Truth-Teller (Ex 21)Die is 6 (1/6) vs Not 6 (5/6)Man reports "It is a 6"P(A|E₁) = 3/4, P(A|E₂) = 1/4
🧒 NCERT Example 19 (Factory Bolts — Full 4-Mark Solution):
Machines A, B, C manufacture 25%, 35%, 40% of bolts; defective outputs are 5%, 4%, 2%. One bolt is drawn and found defective (D). Find P(from Machine B | D):
• Priors: P(E₁)=0.25, P(E₂)=0.35, P(E₃)=0.40.  Likelihoods: P(D|E₁)=0.05, P(D|E₂)=0.04, P(D|E₃)=0.02.
• P(E₂ | D) = 0.35 × 0.040.25×0.05 + 0.35×0.04 + 0.40×0.02 = 0.01400.0125 + 0.0140 + 0.0080 = 140345 = 2869!
Watch TWO sneaky traps in 4/5-mark Bayes questions! (1) In Medical Test questions, NEVER assume False Positive rate equals 1 − True Positive rate! Read BOTH sentences separately: e.g., "90% detected in sick" (P(+|E) = 0.90) vs "1% diagnosed positive in healthy" (P(+|E') = 0.01 — NOT 0.10!). (2) In Factory/Urn questions, keep decimal places aligned (0.0125 + 0.0140 + 0.0080 = 0.0345) — misplacing one zero turns 0.0080 into 0.080!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE Model / PYQ • 1 Marks): If P(A) = 0.8, P(B) = 0.5, and P(B | A) = 0.4, find P(A ∩ B) and P(A | B).

P(A ∩ B) = P(A) · P(B | A) = 0.8 × 0.4 = 0.32.

P(A | B) = P(A ∩ B)P(B) = 0.320.5 = 0.64.

Q2 (Kerala HSE March PYQ • 2 Marks): Evaluate P(A ∪ B) if 2P(A) = P(B) = 5/13 and P(A | B) = 2/5.

Here P(A) = 5/26 and P(B) = 5/13 = 10/26.

P(A ∩ B) = P(B) · P(A | B) = (5/13) × (2/5) = 2/13 = 4/26.

P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 5 + 10 − 426 = 1126.

Q3 (Kerala HSE March PYQ • 2 Marks): Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that (i) both are red, (ii) one is black and other is red.

Total balls = 18. With replacement, draws are independent: P(Red) = 8/18 = 4/9, P(Black) = 10/18 = 5/9.

(i) P(both Red) = (4/9) × (4/9) = 16/81.

(ii) P(one Black, one Red) = P(BR) + P(RB) = (5/9)(4/9) + (4/9)(5/9) = 40/81.

Q4 (NCERT / Kerala HSE PYQ • 3 Marks): Probability of solving a specific problem independently by A and B are ½ and ⅓ respectively. If both try independently, find the probability that (i) the problem is solved, (ii) exactly one of them solves the problem.

Since A and B are independent, P(A) = ½ ⇒ P(A') = ½, and P(B) = ⅓ ⇒ P(B') = ⅔.

(i) P(solved) = 1 − P(A')P(B') = 1 − (½)(⅔) = 1 − ⅓ = ⅔.

(ii) P(exactly one) = P(A)P(B') + P(A')P(B) = (½)(⅔) + (½)(⅓) = ⅓ + ⅙ = ½.

Q5 (Kerala HSE Say / Model PYQ • 3 Marks): A die is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?

Let E = '4 appears at least once', F = 'sum of numbers is 6'.

F = {(1,5), (2,4), (3,3), (4,2), (5,1)} → n(F) = 5.

E ∩ F = {(2,4), (4,2)} → n(E ∩ F) = 2.

P(E | F) = n(E ∩ F)n(F) = 25.

Q6 (Kerala HSE March PYQ (High Yield) • 4 Marks): Bag I contains 3 red and 4 black balls while Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from Bag II.

Let E₁ = Bag I chosen, E₂ = Bag II chosen (P(E₁) = P(E₂) = ½); A = red ball drawn.

P(A | E₁) = 3/7,  P(A | E₂) = 5/11.

By Bayes' Theorem:

P(E₂ | A) = P(E₂)P(A|E₂)P(E₁)P(A|E₁) + P(E₂)P(A|E₂) = (½)(5/11)(½)(3/7) + (½)(5/11) = 5/11(33 + 35)/77 = 3568.

Q7 (NCERT Ex 19 / HSE PYQ • 4 Marks): In a factory manufacturing bolts, machines A, B, and C manufacture respectively 25%, 35%, and 40% of the bolts. Of their outputs, 5%, 4%, and 2% are defective. A bolt drawn at random is found to be defective. Find the probability that it was manufactured by machine B.

Let B₁, B₂, B₃ = bolt made by A, B, C; E = defective bolt.

P(B₁)=0.25, P(B₂)=0.35, P(B₃)=0.40; P(E|B₁)=0.05, P(E|B₂)=0.04, P(E|B₃)=0.02.

P(B₂ | E) = 0.35 × 0.040.25×0.05 + 0.35×0.04 + 0.40×0.02 = 0.01400.0125 + 0.0140 + 0.0080 = 0.01400.0345 = 2869.

Q8 (NCERT Ex 18 / HSE Model PYQ • 4 Marks): In a population, 0.1% actually have a disease. A test detects 90% of diseased people correctly, while 1% of healthy people wrongly test positive. If a random person tests positive, what is the probability they actually have the disease?

Let E = person has disease (P(E) = 0.001, P(E') = 0.999); A = test is positive.

P(A | E) = 0.90,  P(A | E') = 0.01.

P(E | A) = 0.001 × 0.900.001 × 0.90 + 0.999 × 0.01 = 0.000900.00090 + 0.00999 = 901089 = 10121 ≈ 0.083.

Q9 (Kerala HSE Say / NCERT Ex 20 PYQ • 4 Marks): A doctor visits a patient by train, bus, scooter, or other means with probabilities 3/10, 1/5, 1/10, 2/5. Probabilities of being late are 1/4, 1/3, 1/12 for train, bus, scooter, and 0 otherwise. When he arrives, he is late. Find the probability he came by train.

Priors: P(T₁)=3/10, P(T₂)=1/5, P(T₃)=1/10, P(T₄)=2/5.

Late likelihoods: P(L|T₁)=1/4, P(L|T₂)=1/3, P(L|T₃)=1/12, P(L|T₄)=0.

P(L) = (3/10)(1/4) + (1/5)(1/3) + (1/10)(1/12) + 0 = 340 + 115 + 1120 = 9 + 8 + 1120 = 18120 = 320.

P(T₁ | L) = 3/4018/120 = 918 = 12!