HOTS Questions on Class 10 Maths Chapter 14 ‘Probability’ with Answers

HOTS Questions on Class 10 Maths Chapter 14 Probability help students understand chance, outcomes, and events in a simple way. They improve knowledge of sample space, probability formulas, complementary events, and real life situations involving dice, coins, and cards. This chapter builds a strong base for solving probability problems and improving math skills. A downloadable PDF is also available for easy revision and practice.

HOTS Questions Based on Theoretical Probability

Question 1: A class has 30 students. 18 play cricket, 12 play football, and 6 play both. A student is selected at random. Find the probability that the student: a) Plays cricket only b) Plays football only c) Plays neither sport d) Plays at least one sport

Answer:

Using set theory:

Cricket only = 18 - 6 = 12

Football only = 12 - 6 = 6

Both = 6

Neither = 30 - (12 + 6 + 6) = 30 - 24 = 6

[Cricket: 12 | Both: 6 | Football: 6] [Neither: 6]

a) P(Cricket only) = 12/30 = 2/5

b) P(Football only) = 6/30 = 1/5

c) P(Neither) = 6/30 = 1/5

d) P(At least one sport):

= P(Cricket + Football + Both) / Total

= 24/30 = 4/5

OR

P(At least one) = 1 - P(Neither) = 1 - 1/5 = 4/5

Answers: a) 2/5, b) 1/5, c) 1/5, d) 4/5

Question 2: Cards numbered 1 to 50 are placed in a box and one is drawn at random. Find the probability that the number drawn is: a) A prime number b) A perfect square c) Divisible by both 3 and 5 d) Neither prime nor composite

Answer:

n(S) = 50

a) Prime numbers from 1-50:

2,3,5,7,11,13,17,19,23,29,31,37,41,43,47

Count = 15

P(prime) = 15/50 = 3/10

b) Perfect squares from 1-50:

1, 4, 9, 16, 25, 36, 49

Count = 7

P(perfect square) = 7/50

c) Divisible by both 3 AND 5 = divisible by 15:

15, 30, 45

Count = 3

P(div by 3 and 5) = 3/50

d) Neither prime nor composite:

Only number that is neither = 1

Count = 1

P = 1/50

Answers: a) 3/10, b) 7/50, c) 3/50, d) 1/50

Question 3: Two different dice are thrown together. Find the probability that the sum of the numbers appearing is: a) Less than 6 b) A perfect square c) A prime number d) Equal to 7

Answer:

n(S) = 36 (6 × 6 combinations)

a) Sum less than 6 (sum = 2,3,4,5):

Sum 2: (1,1) → 1 way

Sum 3: (1,2)(2,1) → 2 ways

Sum 4: (1,3)(2,2)(3,1) → 3 ways

Sum 5: (1,4)(2,3)(3,2)(4,1) → 4 ways

Total = 1+2+3+4 = 10

P = 10/36 = 5/18

b) Perfect square sums (4, 9):

Sum 4: 3 ways

Sum 9: (3,6)(4,5)(5,4)(6,3) → 4 ways

Total = 7

P = 7/36

c) Prime sums (2,3,5,7,11):

Sum 2: 1 way

Sum 3: 2 ways

Sum 5: 4 ways

Sum 7: (1,6)(2,5)(3,4)(4,3)(5,2)(6,1) → 6 ways

Sum 11: (5,6)(6,5) → 2 ways

Total = 1+2+4+6+2 = 15

P = 15/36 = 5/12

d) Sum equal to 7:

(1,6)(2,5)(3,4)(4,3)(5,2)(6,1) → 6 ways

P = 6/36 = 1/6

Answers: a) 5/18, b) 7/36, c) 5/12, d) 1/6

Question 4: A bag contains x red balls and y blue balls. One ball is drawn at random. The probability of drawing a red ball is 2/5. If 4 more blue balls are added, the probability of drawing a red ball becomes 1/3. Find the original number of red and blue balls.

Answer:

Initially:

P(red) = x/(x+y) = 2/5

5x = 2(x+y)

5x = 2x + 2y

3x = 2y

y = 3x/2  ...(1)

After adding 4 blue balls:

P(red) = x/(x+y+4) = 1/3

3x = x + y + 4

2x = y + 4

2x - y = 4  ...(2)

Substituting (1) in (2):

2x - 3x/2 = 4

4x/2 - 3x/2 = 4

x/2 = 4

x = 8

y = 3 × 8/2 = 12

Verification:

Initial: P(red) = 8/20 = 2/5

After adding 4 blue: P(red) = 8/24 = 1/3

Answer: Red balls = 8, Blue balls = 12

HOTS Questions Based on Experimental Probability

 

Question 5: A thumbtack is tossed 200 times. It lands with the point up 130 times and point down 70 times.

a) Find the experimental probability of point up. b) Find experimental probability of point down. c) Are both outcomes equally likely? Explain. d) If the tack is tossed 1000 times, how many times would you expect point up?

Answer Explanation:

Total trials = 200

Point up = 130

Point down = 70

a) P(point up) = 130/200 = 13/20 = 0.65

b) P(point down) = 70/200 = 7/20 = 0.35

c) No, they are NOT equally likely.

If equally likely, each probability = 1/2 = 0.5

But P(up) = 0.65 ≠ 0.50

The shape of the tack means it naturally

lands point up more often. This is NOT

like a fair coin physical shape matters.

d) Expected point up in 1000 tosses:

= P(point up) × total trials

= 0.65 × 1000

= 650 times

Answers: a) 13/20, b) 7/20, c) No, d) 650 times

Question 6: A die is rolled 120 times. The frequency table is:

a) Find experimental probability for each outcome. b) Find theoretical probability for each outcome. c) Compare and comment on whether the die appears fair. d) How many times would you expect 6 in 600 rolls?

Answer Explanation:

Total rolls = 120

Theoretical probability for each = 1/6 ≈ 0.167

a) Experimental probabilities:

P(1) = 20/120 = 1/6

P(2) = 22/120 = 11/60

P(3) = 18/120 = 3/20

P(4) = 21/120 = 7/40

P(5) = 19/120

P(6) = 20/120 = 1/6

b) Theoretical P(any number) = 1/6 for all

c) Comparison:

All experimental probabilities are very close

to theoretical probability (1/6 ≈ 0.167).

Maximum deviation is P(2) = 0.183 vs 0.167.

This small variation is normal for 120 trials.

The die APPEARS to be fair.

As number of trials increases, experimental

probability approaches theoretical probability.

(Law of Large Numbers)

d) Expected 6 in 600 rolls:

= (1/6) × 600 = 100 times

HOTS Questions Based on Dice, Coins and Cards

Question 7: Two coins are tossed simultaneously 400 times. The results are: Two heads: 100 times One head, one tail: 180 times Two tails: 120 times

a) Find experimental probability for each outcome. b) Compare with theoretical probabilities. c) Are the coins fair based on experimental results?

Answer:

Total = 400

a) Experimental:

P(HH) = 100/400 = 1/4

P(HT or TH) = 180/400 = 9/20

P(TT) = 120/400 = 3/10

b) Theoretical (fair coins):

Sample space = {HH, HT, TH, TT}

P(HH) = 1/4

P(one head) = 2/4 = 1/2

P(TT) = 1/4

c) Comparison:

P(HH): Experimental 1/4, Theoretical 1/4

P(one head): Experimental 9/20 = 0.45

             Theoretical 1/2 = 0.50 (slight difference)

P(TT): Experimental 3/10 = 0.30

       Theoretical 1/4 = 0.25 (slight difference)

The coins may have slight bias but with only

400 trials, this variation is somewhat normal.

More trials needed to confirm fairness.

Question 8: A card is drawn from a well shuffled deck of 52 cards. Find the probability that the card is: a) A king or a queen b) A red card or a king c) Neither a heart nor a king d) A face card but not a king

Answer:

n(S) = 52

a) Kings = 4, Queens = 4 (mutually exclusive)

P(king or queen) = (4+4)/52 = 8/52 = 2/13

b) Red cards = 26, Kings = 4, Red kings = 2

P(red or king) = (26 + 4 - 2)/52 = 28/52 = 7/13

c) Hearts = 13, Kings = 4, King of hearts = 1

Cards that ARE hearts or kings = 13 + 4 - 1 = 16

Cards that are NEITHER = 52 - 16 = 36

P = 36/52 = 9/13

d) Face cards = 3 per suit × 4 = 12

(Jacks, Queens, Kings)

Face cards that are NOT kings = 12 - 4 = 8

(Jacks + Queens = 4 + 4 = 8)

P = 8/52 = 2/13

Answers: a) 2/13, b) 7/13, c) 9/13, d) 2/13

Question 9: Three coins are tossed together. Find the probability of getting: a) At least two heads b) Exactly one tail c) At most two tails d) All same faces

Answer:

Sample space for 3 coins:

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

n(S) = 8

a) At least two heads (HH or HHH):

{HHH, HHT, HTH, THH} → 4 outcomes

P = 4/8 = 1/2

b) Exactly one tail:

{HHT, HTH, THH} → 3 outcomes

P = 3/8

c) At most two tails (0, 1, or 2 tails):

Complement = all three tails = {TTT} → 1

P(at most 2 tails) = 1 - P(TTT) = 1 - 1/8 = 7/8

d) All same faces {HHH, TTT} → 2 outcomes

P = 2/8 = 1/4

Answers: a) 1/2, b) 3/8, c) 7/8, d) 1/4

HOTS Questions Based on Real-Life Applications

Question 10: In a lucky draw, 100 tickets are sold. First prize: 1 ticket, Second prize: 2 tickets, Third prize: 5 tickets. Rohan buys 3 tickets.

a) Probability Rohan wins the first prize. b) Probability Rohan wins at least one prize. c) Probability Rohan wins no prize. d) If tickets cost ₹10 each and prizes are ₹500, ₹200, ₹100, find Rohan's expected return.

Answer:

Total tickets = 100

Rohan's tickets = 3

a) P(first prize) = 3/100

(Rohan has 3 chances for 1 winning ticket)

Wait: P(Rohan wins first) = 3/100

(any of his 3 tickets could be the winner)

b) Prize tickets = 1 + 2 + 5 = 8

P(Rohan wins at least one prize):

P = 1 - P(no prize)

= 1 - (97/100 × 96/99 × 95/98)

Approximate method:

P(no prize) ≈ (97/100)³ ≈ 0.912

P(at least one prize) ≈ 1 - 0.912 = 0.088

Simple approximation:

Winning tickets = 8, Non-winning = 92

P(Rohan wins) ≈ 3 × 8/100 = 24/100 = 6/25

(Using simple proportion)

c) P(no prize) ≈ 1 - 6/25 = 19/25

d) Expected return:

P(₹500) = 3/100, P(₹200) = 6/100, P(₹100) = 15/100

Expected = 500×(3/100) + 200×(6/100) + 100×(15/100)

= 15 + 12 + 15 = ₹42

Cost = 3 × ₹10 = ₹30

Expected profit = 42 - 30 = ₹12

Answers: a) 3/100, b) ≈ 6/25, c) ≈ 19/25, d) Expected return ₹42

Question 11: The weather forecast says the probability of rain on Saturday is 0.6 and on Sunday is 0.4. These events are independent.

a) Probability it rains on both days. b) Probability it rains on at least one day. c) Probability it rains on neither day. d) Probability it rains on exactly one day.

Answer Explanation:

P(Rain Saturday) = 0.6

P(Rain Sunday) = 0.4

Events are independent.

a) P(Rain both days):

= P(Sat) × P(Sun)

= 0.6 × 0.4

= 0.24

b) P(At least one day):

= 1 - P(no rain both days)

= 1 - P(no rain Sat) × P(no rain Sun)

= 1 - 0.4 × 0.6

= 1 - 0.24

= 0.76

c) P(Rain neither day):

= P(no Sat) × P(no Sun)

= 0.4 × 0.6

= 0.24

d) P(Exactly one day):

= P(Sat only) + P(Sun only)

= P(Sat) × P(no Sun) + P(no Sat) × P(Sun)

= 0.6 × 0.6 + 0.4 × 0.4

= 0.36 + 0.16

= 0.52

Answers: a) 0.24, b) 0.76, c) 0.24, d) 0.52

Question 12: A factory produces 1000 light bulbs per day. Quality control shows that 50 bulbs are defective. A bulb is selected at random.

a) Find P(defective). b) Find P(non-defective). c) If a shopkeeper buys 20 bulbs, how many defective ones would you expect? d) A customer claims the bulb is not defective. What is the probability the customer is correct?

Answer:

Total = 1000, Defective = 50, Non-defective = 950

a) P(defective) = 50/1000 = 1/20 = 0.05

b) P(non-defective) = 950/1000 = 19/20 = 0.95

c) Expected defective in 20 bulbs:

= P(defective) × 20

= (1/20) × 20

= 1 bulb expected

d) P(customer is correct) = P(non-defective)

= 19/20 = 0.95

The customer has a 95% chance of being correct.

Answers: a) 1/20, b) 19/20, c) 1 bulb, d) 19/20 or 95%

HOTS Questions Based on Data Interpretation

HOTS Question 13: The blood groups of 40 students in a class are recorded:

A student is selected randomly. Find: a) P(blood group B) b) P(blood group AB or O) c) P(blood group not A) d) Which blood group has the highest probability?

Answer Explanation:

Total = 40

a) P(B) = 10/40 = 1/4

b) P(AB or O) = (6+12)/40 = 18/40 = 9/20

c) P(not A) = (40-12)/40 = 28/40 = 7/10

d) Blood groups A and O both have

frequency 12 both have P = 12/40 = 3/10

A and O are equally highest probability.

Answers: a) 1/4, b) 9/20, c) 7/10, d) A and O (tied at 3/10)

Question 14: 500 families were surveyed about number of children:

A family is chosen randomly. Find: a) P(family has exactly 2 children) b) P(family has at most 2 children) c) P(family has more than 2 children) d) P(family has at least 1 child)

Answer Explanation:

Total = 500

a) P(exactly 2) = 200/500 = 2/5

b) P(at most 2) = P(0 or 1 or 2)

= (50 + 150 + 200)/500

= 400/500 = 4/5

c) P(more than 2) = P(3 or 4+)

= (75 + 25)/500

= 100/500 = 1/5

Verify: 4/5 + 1/5 = 1 (complement check)

d) P(at least 1 child) = 1 - P(0 children)

= 1 - 50/500

= 1 - 1/10

= 9/10

Answers: a) 2/5, b) 4/5, c) 1/5, d) 9/10

HOTS Questions Involving Logical Reasoning

Question 15: Two students find P(getting a sum of 6 when two dice are rolled).

Student A says: Possible sums = 2 to 12 = 11 values. P(sum = 6) = 1/11

Student B says: Total outcomes = 36. Sum 6 occurs as (1,5)(2,4)(3,3)(4,2)(5,1) = 5 ways. P(sum = 6) = 5/36

Who is correct? Explain the error in the wrong approach.

Answer:

Student B is CORRECT.

Error in Student A's reasoning:

Student A assumed all sums from 2 to 12

are equally likely. This is WRONG.

Why are they not equally likely?

Sum 2 can only happen 1 way: (1,1)

Sum 7 can happen 6 ways: (1,6)(2,5)...

They have very different frequencies!

The sample space has 36 equally likely outcomes,

not 11 equally likely sums.

Correct approach (Student B):

Always work with the FUNDAMENTAL sample space

of individual outcomes (36), not derived values.

P(sum = 6) = 5/36

Answer: Student B is correct. The 11 sum values are NOT equally likely.

HOTS Question 16: A student solved: "Find P(drawing an ace or a spade from a deck of 52 cards)."

Student's work:

P(ace) = 4/52

P(spade) = 13/52

P(ace or spade) = 4/52 + 13/52 = 17/52

Find the error and give the correct answer.

Answer:

The student made an error by NOT accounting

for the overlap (ace of spades appears in BOTH).

This is the inclusion-exclusion principle error.

Aces and spades are NOT mutually exclusive

because the Ace of Spades belongs to BOTH groups.

Correct formula:

P(A or B) = P(A) + P(B) - P(A and B)

P(ace or spade) = P(ace) + P(spade) - P(ace of spades)

= 4/52 + 13/52 - 1/52

= 16/52

= 4/13

The student over-counted the Ace of Spades

by including it in BOTH groups.

Correct answer = 4/13, not 17/52.

Correct Answer: P(ace or spade) = 4/13

Question 17: Find P(at least one head when 3 coins are tossed) using: Method 1: Direct counting Method 2: Complement method

Which method is faster? When is complement method better?

Answer:

Method 1: Direct counting

Favorable outcomes (at least 1 head):

{HHH, HHT, HTH, THH, HTT, THT, TTH} = 7

P = 7/8

Method 2: Complement

P(at least 1 head) = 1 - P(no head)

P(all tails) = P(TTT) = 1/8

P(at least 1 head) = 1 - 1/8 = 7/8

Both give 7/8.

Method 2 is FASTER here because:

"No head" has only 1 outcome (TTT)

"At least 1 head" has 7 outcomes (tedious to list)

When to use complement method:

• When "at least" or "at most" appears
• When direct counting needs many outcomes
• When the complement has very few outcomes

When direct method works better:

• Small sample spaces
• When complement is complex
• When you need to list all favorable outcomes anyway

Answer: Both give 7/8. Complement method is faster.

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