Chapter 14 : Class 10 : Probability : Case Study Question Bank

✦ NCF 2023 • COMPETENCY-FOCUSED QUESTION BANK ✦

Chapter 14: Probability

Class 10 Standard Mathematics (Case Study Bank)

Curated By Akash Srivastva

✨ Interactive TLMs, Worksheets & Digital Study Notes: www.akashmaths.online

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Case Study 1 Game Theory 4 Marks

Digital Board Game & Rolling Dice

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Two friends play a digital board game where token advances depend on simultaneously rolling two fair six-sided dice (each numbered 1 to 6). Total outcomes in sample space \(N(S) = 36\).

Q1. Prime Sum Probability 1 Mark

Find the probability that the sum of the numbers on the top faces of the two dice is a prime number.

(A) 1/3

(B) 5/12

(C) 7/12

(D) 1/2

Q2. Square Product Probability 1 Mark

Find the probability that the product of the numbers on the two dice is a perfect square.

(A) 2/9

(B) 1/4

(C) 1/6

(D) 5/18

Q3. Difference & Doublet Outcomes 2 Marks

Find the probability that the absolute difference between numbers on the two dice is exactly 2.

(A) 1/4

(B) 5/18

(C) 2/9

(D) 7/36

OR (Alternative Q3)

Find the probability that the roll results in a "doublet of even numbers".

(A) 1/12

(B) 1/6

(C) 1/4

(D) 1/18

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	Total outcomes = 36. Prime sums {2, 3, 5, 7, 11} have 1 + 2 + 4 + 6 + 2 = 15 outcomes. P(Prime Sum) = 15/36 = 5/12.	[0.5 Mark] for identifying 15 outcomes. [0.5 Mark] for 5/12.
Q2	Products forming perfect squares: 1:(1,1), 4:(1,4),(2,2),(4,1), 9:(3,3), 16:(4,4), 25:(5,5), 36:(6,6) ⇒ 8 outcomes. P(Square Product) = 8/36 = 2/9.	[0.5 Mark] for listing 8 outcomes. [0.5 Mark] for 2/9.
Q3	|x − y| = 2 outcomes: (1,3), (3,1), (2,4), (4,2), (3,5), (5,3), (4,6), (6,4) ⇒ 8 outcomes. P(|x − y| = 2) = 8/36 = 2/9.	[1 Mark] for listing 8 pairs. [1 Mark] for 2/9.
Q3 (OR)	Even doublets: (2,2), (4,4), (6,6) ⇒ 3 outcomes. P(Even Doublet) = 3/36 = 1/12.	[1 Mark] for listing 3 pairs. [1 Mark] for 1/12.

Case Study 2 Quality Control 4 Marks

Microchip Manufacturing & Defect Rate

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A batch of 48 microchips contains x defective microchips. If 12 more defective microchips are added, probability of drawing a defective chip doubles.

Q1. Finding Defective Chips (x) 1 Mark

Formulate an algebraic equation and calculate the exact value of x in the initial batch.

(A) x = 6

(B) x = 8

(C) x = 10

(D) x = 12

Q2. Initial Perfect Chip Probability 1 Mark

Calculate the probability of drawing a non-defective chip from the initial batch of 48.

(A) 5/6

(B) 3/4

(C) 7/8

(D) 2/3

Q3. Batch Addition & Removal 2 Marks

From the new batch (after adding 12 defective chips), find probability of drawing a non-defective chip.

(A) 1/2

(B) 3/4

(C) 2/3

(D) 5/6

OR (Alternative Q3)

If 6 non-defective chips are removed from the initial batch of 48, find revised probability of defective chip.

(A) 4/21

(B) 2/21

(C) 5/21

(D) 1/7

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	(x + 12)/60 = 2(x/48) = x/24 ⇒ 24(x + 12) = 60x ⇒ 36x = 288 ⇒ x = 8.	[0.5 Mark] for setting equation. [0.5 Mark] for solving x = 8.
Q2	Non-defective chips = 48 − 8 = 40 ⇒ P(Non-defective) = 40/48 = 5/6.	[1 Mark] for 5/6.
Q3	New total = 48 + 12 = 60. Non-defective = 40. P(Non-defective in new batch) = 40/60 = 2/3.	[1 Mark] for 40/60. [1 Mark] for 2/3.
Q3 (OR)	Remaining total = 48 − 6 = 42 chips. Defective chips = 8. P(Defective) = 8/42 = 4/21.	[1 Mark] for remaining count 42. [1 Mark] for 4/21.

Case Study 3 Meteorology AI 4 Marks

AI Predictive Meteorology

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An AI predictive weather model calculates probability of heavy precipitation as \(P(R) = \frac{2x - 3}{10}\) and no precipitation as \(P(R') = \frac{3x - 2}{10}\).

Q1. Complementary Value (x) 1 Mark

Using the relationship P(R) + P(R') = 1, find the value of algebraic variable x.

(A) x = 2

(B) x = 3

(C) x = 4

(D) x = 5

Q2. Precipitation Probability 1 Mark

Calculate the exact numerical probability of precipitation P(R).

(A) 0.3 (or 3/10)

(B) 0.4

(C) 0.5

(D) 0.2

Q3. Union Probability & Forecast Horizon 2 Marks

If P(High Wind) = 0.45 and P(Precipitation ∩ Wind) = 0.15, find P(Precipitation ∪ Wind).

(A) 0.50

(B) 0.75

(C) 0.60 (or 3/5)

(D) 0.45

OR (Alternative Q3)

Find probability of no rain if P(Rain) = 0.3, and revised no-rain probability if P(Rain) becomes 2/5.

(A) 0.7 and 0.6 (or 3/5)

(B) 0.6 and 0.5

(C) 0.8 and 0.6

(D) 0.7 and 0.5

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	P(R) + P(R') = 1 ⇒ (2x − 3)/10 + (3x − 2)/10 = 1 ⇒ 5x − 5 = 10 ⇒ 5x = 15 ⇒ x = 3.	[0.5 Mark] for complementary relation. [0.5 Mark] for x = 3.
Q2	P(R) = [2(3) − 3]/10 = 3/10 = 0.3.	[1 Mark] for P(R) = 0.3.
Q3	P(R ∪ W) = P(R) + P(W) − P(R ∩ W) = 0.3 + 0.45 − 0.15 = 0.6.	[1 Mark] for addition rule formula. [1 Mark] for 0.6.
Q3 (OR)	P(No Rain) = 1 − 0.3 = 0.7. Revised P(No Rain) = 1 − 2/5 = 3/5 = 0.6.	[1 Mark] for 0.7. [1 Mark] for 0.6.

Case Study 4 Loyalty Gaming 4 Marks

Lucky Spin Reward Wheel

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An e-commerce digital reward wheel is divided into 12 equal sectors numbered 1 to 12. An arrow pointer randomly selects a sector upon deceleration.

Q1. Even Number Probability 1 Mark

Find the probability that the spinning wheel stops on an even-numbered sector.

(A) 1/3

(B) 1/2

(C) 5/12

(D) 7/12

Q2. Multiple of 3 Probability 1 Mark

Find the probability that the wheel stops on a sector whose number is a multiple of 3.

(A) 1/3

(B) 1/4

(C) 1/2

(D) 5/12

Q3. Prime Number & Compound Events 2 Marks

Find the probability that the wheel stops on a prime-numbered sector.

(A) 1/2

(B) 1/3

(C) 5/12

(D) 7/12

OR (Alternative Q3)

Find the probability that the wheel stops on a sector that is a multiple of 4 or a multiple of 5.

(A) 5/12

(B) 1/2

(C) 1/3

(D) 1/4

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	Total outcomes = 12. Even numbers: {2, 4, 6, 8, 10, 12} ⇒ 6 outcomes. P(Even) = 6/12 = 1/2.	[0.5 Mark] for even count. [0.5 Mark] for 1/2.
Q2	Multiples of 3: {3, 6, 9, 12} ⇒ 4 outcomes. P(Multiple of 3) = 4/12 = 1/3.	[1 Mark] for 1/3.
Q3	Prime numbers in 1 to 12: {2, 3, 5, 7, 11} ⇒ 5 outcomes. P(Prime) = 5/12.	[1 Mark] for identifying primes. [1 Mark] for 5/12.
Q3 (OR)	Multiples of 4: {4, 8, 12}; Multiples of 5: {5, 10} ⇒ Union: {4, 5, 8, 10, 12} (5 outcomes). P(Multiple of 4 or 5) = 5/12.	[1 Mark] for identifying 5 outcomes. [1 Mark] for 5/12.

Case Study 5 Green Transport 4 Marks

Eco-Friendly School Travel

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A school survey records daily commuting modes of 120 Class 10 students:

Commuting Mode	Students
Electric Bus (EB)	45
Bicycle (BI)	35
Walking (WA)	25
Private Car (PC)	15

Q1. Active Mode Probability 1 Mark

Find the probability that a randomly selected student commutes by Bicycle or Walking.

(A) 5/12

(B) 1/2 (or 0.5)

(C) 7/12

(D) 1/3

Q2. Complementary Probability 1 Mark

Calculate the probability that the selected student does NOT commute by Private Car.

(A) 7/8 (or 0.875)

(B) 5/8

(C) 3/4

(D) 11/12

Q3. Sample Modification & Alternative Union 2 Marks

If 5 Bicycle and 5 Electric Bus commuters join, find new probability of Electric Bus.

(A) 5/12

(B) 6/13

(C) 5/13

(D) 1/2

OR (Alternative Q3)

For the original group of 120, find probability of commuting by Electric Bus or Private Car.

(A) 1/2 (or 0.5)

(B) 5/12

(C) 7/12

(D) 3/8

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	Favourable (Bicycle + Walking) = 35 + 25 = 60. P(Active Mode) = 60/120 = 1/2 = 0.5.	[0.5 Mark] for favourable count. [0.5 Mark] for 1/2.
Q2	P(PC) = 15/120 = 1/8 ⇒ P(NOT PC) = 1 − 1/8 = 7/8 = 0.875.	[1 Mark] for 7/8.
Q3	New total = 120 + 10 = 130. New Electric Bus commuters = 45 + 5 = 50. P(EB_new) = 50/130 = 5/13.	[1 Mark] for new counts. [1 Mark] for 5/13.
Q3 (OR)	Favourable (EB or PC) = 45 + 15 = 60. P(EB or PC) = 60/120 = 1/2 = 0.5.	[1 Mark] for favourable count 60. [1 Mark] for 1/2.

Case Study 6 Conservation Biology 4 Marks

Wetland Fish Population Study

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A wetland research sample netting contains 80 fish: Male Rohu: 18, Female Rohu: 22, Male Katla: 12, Female Katla: 28. One fish is selected at random.

Q1. Female Fish Probability 1 Mark

Find the probability that the selected fish is a female fish of either species.

(A) 1/2

(B) 5/8 (or 0.625)

(C) 3/4

(D) 7/10

Q2. Species Identification 1 Mark

Find the probability that the selected fish belongs to the Rohu species.

(A) 1/2 (or 0.5)

(B) 3/8

(C) 5/8

(D) 2/5

Q3. Species Replenishment & Union 2 Marks

If Male Katla fish are added so P(Male Katla) = 1/5, calculate how many were added.

(A) 4 fish

(B) 6 fish

(C) 5 fish

(D) 8 fish

OR (Alternative Q3)

From original group, find probability of selecting Male Rohu or Female Katla.

(A) 23/40 (or 0.575)

(B) 21/40

(C) 1/2

(D) 11/20

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	Total fish = 18 + 22 + 12 + 28 = 80. Females = 22 + 28 = 50. P(Female) = 50/80 = 5/8.	[0.5 Mark] for favourable count 50. [0.5 Mark] for 5/8.
Q2	Total Rohu = 18 + 22 = 40. P(Rohu) = 40/80 = 1/2.	[1 Mark] for 1/2.
Q3	(12 + y)/(80 + y) = 1/5 ⇒ 5(12 + y) = 80 + y ⇒ 60 + 5y = 80 + y ⇒ 4y = 20 ⇒ y = 5 fish.	[1 Mark] for algebraic formulation. [1 Mark] for y = 5 fish.
Q3 (OR)	Favourable (Male Rohu + Female Katla) = 18 + 28 = 46. P = 46/80 = 23/40 = 0.575.	[1 Mark] for count 46. [1 Mark] for 23/40.

Case Study 7 Algorithmic Gaming 4 Marks

Digital Deck of Cards Analysis

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A card-shuffling simulation engine for an online board game operates with a standard, complete deck of 52 playing cards.

Q1. Red Face Card Probability 1 Mark

Find the probability that a single card drawn at random is a red face card (King, Queen, Jack).

(A) 3/13

(B) 3/26

(C) 1/13

(D) 6/13

Q2. Spade or Ace Probability 1 Mark

Find the probability that the drawn card is either a spade or an ace.

(A) 4/13

(B) 1/4

(C) 17/52

(D) 5/13

Q3. Black Royal Card Removal 2 Marks

If all black queens and black kings are removed, find P(face card) and P(red card).

(A) (i) 1/4, (ii) 1/2

(B) (i) 1/6, (ii) 1/2

(C) (i) 1/6, (ii) 13/24

(D) (i) 1/8, (ii) 13/24

OR (Alternative Q3)

For a full deck of 52 cards, find the probability that drawn card is neither a king nor a red queen.

(A) 23/26

(B) 11/13

(C) 21/26

(D) 25/26

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	Total red face cards = 6 (J, Q, K of Hearts & Diamonds). P = 6/52 = 3/26.	[1 Mark] for 3/26.
Q2	Spades = 13, other Aces = 3 ⇒ Favourable = 16. P = 16/52 = 4/13.	[1 Mark] for 4/13.
Q3	Cards removed = 4 ⇒ Total remaining = 48. Remaining face cards = 8 ⇒ P(Face) = 8/48 = 1/6. Red cards remaining = 26 ⇒ P(Red) = 26/48 = 13/24.	[1 Mark] for (i) 1/6. [1 Mark] for (ii) 13/24.
Q3 (OR)	Undesirable cards: 4 Kings + 2 Red Queens = 6. Favourable = 52 − 6 = 46. P(Neither King nor Red Queen) = 46/52 = 23/26.	[1 Mark] for count 46. [1 Mark] for 23/26.

Case Study 8 Sustainable Production 4 Marks

Eco-Friendly Marbles Bag

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A drawstring bag contains clay marbles: x Green, 10 Blue, and 14 Red. A child draws one marble at random.

Q1. Finding Green Marbles (x) 1 Mark

If probability of drawing a green marble is 1/3, calculate the value of x.

(A) x = 10

(B) x = 12

(C) x = 14

(D) x = 16

Q2. Complementary Event 1 Mark

Using x = 12, find the probability that a marble drawn at random is NOT blue.

(A) 13/18

(B) 5/18

(C) 7/9

(D) 2/3

Q3. Bag Reconfiguration & Union 2 Marks

If 2 green are removed and 2 red added, find revised P(Green) and P(Red).

(A) (i) 1/3, (ii) 1/2

(B) (i) 5/18, (ii) 7/18

(C) (i) 5/18, (ii) 4/9

(D) (i) 1/4, (ii) 4/9

OR (Alternative Q3)

For the original bag (x = 12), calculate probability of drawing a blue or a red marble.

(A) 2/3

(B) 3/4

(C) 7/12

(D) 5/6

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	Total = x + 24. x/(x + 24) = 1/3 ⇒ 3x = x + 24 ⇒ 2x = 24 ⇒ x = 12.	[0.5 Mark] for setup. [0.5 Mark] for x = 12.
Q2	Total = 12 + 24 = 36. Not Blue (Green + Red) = 12 + 14 = 26 ⇒ P = 26/36 = 13/18.	[1 Mark] for 13/18.
Q3	New counts: Green = 10, Red = 16, Blue = 10 (Total = 36). (i) P(Green) = 10/36 = 5/18; (ii) P(Red) = 16/36 = 4/9.	[1 Mark] for (i) 5/18. [1 Mark] for (ii) 4/9.
Q3 (OR)	Favourable (Blue + Red) = 10 + 14 = 24. P(Blue or Red) = 24/36 = 2/3.	[1 Mark] for count 24. [1 Mark] for 2/3.

Live Practice: Chapter 14 Probability

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Case Study 1

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