Probability — Class 10 Maths practice questions

Question 1

A school is organising a 'Science Carnival' in which one of the stalls has a two-stage game. In Stage 1, a contestant spins a fair spinner divided into 5 equal sectors numbered 1, 2, 3, 4, and 5. If the spinner lands on a prime number, the contestant advances to Stage 2; otherwise the game ends and the contestant loses.

In Stage 2, the contestant draws one card at random from a set of 10 cards numbered 1 to 10. The contestant wins a prize only if the card drawn shows a perfect square.

Based on the above situation, answer the following questions:

(i) What is the probability that the spinner lands on a prime number in Stage 1? [1 mark]

(ii) Given that the contestant reaches Stage 2, what is the probability of winning a prize? [1 mark]

(iii) A contestant claims: "The probability that I reach Stage 2 AND win a prize is more than 1/4." Is this claim correct? Find the probability of reaching Stage 2 and winning a prize, and justify your answer. [2 marks]

OR

(iii) Two contestants play the game independently. Find the probability that exactly one of them wins a prize (i.e., reaches Stage 2 and picks a perfect-square card). [2 marks]

Accepted Answer

Part (i) [1 mark]

The 5 equally likely outcomes on the spinner are: {1, 2, 3, 4, 5}.

Prime numbers in this set: {2, 3, 5} → 3 favourable outcomes.

∴ P(spinner lands on a prime number) = 3/5

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Part (ii) [1 mark]

Cards numbered 1 to 10; perfect squares in this range: {1, 4, 9} → 3 favourable outcomes out of 10.

∴ P(winning a prize | reaches Stage 2) = 3/10

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Part (iii) — Main [2 marks]

Let A = event that contestant reaches Stage 2, and W = event that contestant wins a prize.

The two stages are independent of each other.

P(A) = 3/5 [from part (i)]

P(W | A) = 3/10 [from part (ii)]

⟹ P(reaches Stage 2 AND wins a prize) = P(A) × P(W | A)

⟹ P(A ∩ W) = (3/5) × (3/10) = 9/50

Now, checking the contestant's claim:

1/4 = 12.5/50, but 9/50 < 12.5/50

∴ 9/50 < 1/4

The contestant's claim is NOT correct. The probability of reaching Stage 2 and winning a prize is 9/50, which is less than 1/4.

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Part (iii) — OR [2 marks]

Let p = probability that a single contestant reaches Stage 2 AND wins a prize.

From above, p = P(A) × P(W | A) = (3/5) × (3/10) = 9/50

∴ Probability that a single contestant does NOT win = 1 − 9/50 = 41/50

The two contestants play independently.

P(exactly one of the two wins)
= P(1st wins and 2nd does not) + P(1st does not and 2nd wins)

⟹ = p × (1 − p) + (1 − p) × p

⟹ = 2 × p × (1 − p)

⟹ = 2 × (9/50) × (41/50)

⟹ = 2 × 369/2500

∴ P(exactly one contestant wins) = 738/2500 = 369/1250

Question 2

A school conducted a survey among 200 students about their preferred mode of studying. The results are shown in the pie chart below (described by central angles):

• Self-study: 126°
• Online videos: 72°
• Coaching class: 90°
• Group study: 54°
• Library: 18°

Based on this data, answer the following questions:

(i) What is the probability that a randomly selected student prefers online videos?

(ii) What is the probability that a randomly selected student prefers neither self-study nor library?

(iii) Two students are selected one after another with replacement. What is the probability that the first student prefers coaching class and the second student prefers group study?

OR

(iii) A student is selected at random. What is the probability that the student prefers either self-study or online videos?

Accepted Answer

Total number of students surveyed = 200. Total central angle = 360°.

Number of students preferring each mode:
• Self-study: (126/360) × 200 = 70
• Online videos: (72/360) × 200 = 40
• Coaching class: (90/360) × 200 = 50
• Group study: (54/360) × 200 = 30
• Library: (18/360) × 200 = 10

Verification: 70 + 40 + 50 + 30 + 10 = 200 ✓

(i) Finding P(student prefers online videos):

P(E) = Number of favourable outcomes / Total number of outcomes

Number of students preferring online videos = 40

∴ P(online videos) = 40/200 = 1/5

(ii) Finding P(student prefers neither self-study nor library):

Number of students preferring self-study = 70
Number of students preferring library = 10

⟹ Number of students preferring self-study or library = 70 + 10 = 80

⟹ Number of students preferring neither self-study nor library = 200 − 80 = 120

∴ P(neither self-study nor library) = 120/200 = 3/5

(iii) [Main Option]

Finding P(1st student prefers coaching class AND 2nd student prefers group study), with replacement:

Since selection is with replacement, both events are independent.

P(1st prefers coaching class) = 50/200 = 1/4

P(2nd prefers group study) = 30/200 = 3/20

By the multiplication rule for independent events:

P(coaching class then group study) = P(coaching class) × P(group study)

⟹ P = 1/4 × 3/20 = 3/80

∴ The required probability = 3/80

[OR Option]

Finding P(student prefers self-study OR online videos):

Number of students preferring self-study = 70
Number of students preferring online videos = 40

Since a student can prefer only one mode, these events are mutually exclusive.

⟹ Number of students preferring self-study or online videos = 70 + 40 = 110

P(self-study or online videos) = 110/200 = 11/20

∴ The required probability = 11/20

Question 3

A school is organising a Science Fair. Students are asked to spin a game wheel to win prizes. The wheel is divided into 8 equal sectors labelled with the following prize categories:

Sector 1: Gold Medal
Sector 2: Silver Medal
Sector 3: Bronze Medal
Sector 4: Certificate
Sector 5: Certificate
Sector 6: Certificate
Sector 7: Participation Token
Sector 8: Participation Token

After the fair, the teacher also prepares a lucky-draw box containing 20 chits. The chits are numbered 1 to 20. One chit is drawn at random.

Based on the above information, answer the following questions:

(i) What is the probability that the wheel stops at a sector labelled 'Certificate'? [1 mark]

(ii) What is the probability that the wheel does NOT stop at a 'Participation Token' sector? [1 mark]

(iii) From the lucky-draw box, find the probability that the number on the drawn chit is either a perfect square OR a prime number.
[2 marks]

OR

(iii) From the lucky-draw box, two events are defined:
Event A: The number on the chit is a multiple of 3.
Event B: The number on the chit is a multiple of 5.
Find the probability that the number on the chit is a multiple of 3 OR a multiple of 5. Also verify that P(A) + P(B) − P(A and B) gives the same result. [2 marks]

Accepted Answer

(i) Total number of equally likely outcomes = 8 (sectors).

Favourable outcomes (Certificate) = 3 (Sectors 4, 5, 6).

∴ P(Certificate) = 3/8

(ii) Number of 'Participation Token' sectors = 2.

⟹ P(Participation Token) = 2/8 = 1/4

∴ P(NOT Participation Token) = 1 − 1/4 = 3/4

(iii) [MAIN OPTION]

Total number of outcomes = 20 (chits numbered 1 to 20).

Perfect squares from 1 to 20: 1, 4, 9, 16 → 4 numbers.

Prime numbers from 1 to 20: 2, 3, 5, 7, 11, 13, 17, 19 → 8 numbers.

Numbers that are BOTH perfect square AND prime: None (1 is not prime; 4, 9, 16 are not prime).

⟹ Favourable outcomes = 4 + 8 − 0 = 12.

∴ P(perfect square OR prime) = 12/20 = 3/5

[OR OPTION]

Total number of outcomes = 20.

Event A: Multiples of 3 from 1–20 → {3, 6, 9, 12, 15, 18} → 6 numbers.

⟹ P(A) = 6/20 = 3/10

Event B: Multiples of 5 from 1–20 → {5, 10, 15, 20} → 4 numbers.

⟹ P(B) = 4/20 = 1/5

A and B (multiples of both 3 and 5, i.e., multiples of 15) → {15} → 1 number.

⟹ P(A and B) = 1/20

Favourable outcomes for A OR B = {3, 5, 6, 9, 10, 12, 15, 18, 20} → 9 numbers.

∴ P(A OR B) = 9/20

Verification: P(A) + P(B) − P(A and B) = 6/20 + 4/20 − 1/20 = 9/20 ✓

∴ The probability that the number is a multiple of 3 OR a multiple of 5 = 9/20, and the addition rule is verified.

Question 4

A school librarian is organising a 'Lucky Dip' reading challenge. She writes one number each on identical slips of paper — the numbers being all integers from 1 to 40 — folds them, and puts them in a box. A student draws one slip at random.

Based on this situation, answer the following:

(i) What is the probability that the number on the slip is a perfect square?

(ii) What is the probability that the number is divisible by both 3 and 5?

(iii) The librarian says: 'If you draw a prime number greater than 20, you win a prize.' Find the probability of winning a prize.

OR

(iii) The librarian adds a rule: 'If you draw a number that is either a multiple of 7 OR a perfect cube, you get bonus points.' Find the probability of getting bonus points.

Accepted Answer

Total number of slips = 40
∴ Total number of equally likely outcomes = 40

─────────────────────────────────────
Part (i) [1 mark]
─────────────────────────────────────
Perfect squares from 1 to 40:
1, 4, 9, 16, 25, 36
∴ Number of favourable outcomes = 6

P(perfect square) = 6/40 = 3/20

∴ P(number is a perfect square) = 3/20

─────────────────────────────────────
Part (ii) [1 mark]
─────────────────────────────────────
A number divisible by both 3 and 5 is divisible by LCM(3, 5) = 15.
Multiples of 15 from 1 to 40: 15, 30
∴ Number of favourable outcomes = 2

P(divisible by both 3 and 5) = 2/40 = 1/20

∴ P(number is divisible by both 3 and 5) = 1/20

─────────────────────────────────────
Part (iii) — MAIN [2 marks]
─────────────────────────────────────
Prime numbers greater than 20 and ≤ 40:
23, 29, 31, 37
∴ Number of favourable outcomes = 4

P(winning a prize) = 4/40 = 1/10

∴ P(drawing a prime number greater than 20) = 1/10

─────────────────────────────────────
Part (iii) — OR [2 marks]
─────────────────────────────────────
Multiples of 7 from 1 to 40: 7, 14, 21, 28, 35
∴ Number of multiples of 7 = 5

Perfect cubes from 1 to 40: 1, 8, 27
∴ Number of perfect cubes = 3

Numbers that are both a multiple of 7 AND a perfect cube from 1 to 40:
Checking: 7×1=7 (not a perfect cube), 7×8=56>40; 1, 8, 27 are not multiples of 7.
∴ No number from 1 to 40 is both a multiple of 7 and a perfect cube.
∴ The two events are mutually exclusive.

By the addition rule of probability:
Favourable outcomes = 5 + 3 = 8

P(bonus points) = 8/40 = 1/5

∴ P(getting bonus points) = 1/5

Question 5

A school is organising a 'Science Talent Hunt' game for its annual fest. A spinning wheel is divided into 40 equal sectors, numbered 1 to 40. A student spins the wheel once. The rules of the game are:
• If the wheel stops at a PERFECT SQUARE, the student wins a Gold Medal.
• If the wheel stops at a PRIME NUMBER greater than 20, the student wins a Silver Medal.
• If the wheel stops at a number that is a MULTIPLE OF 6 but NOT a multiple of 4, the student wins a Bronze Medal.
• If the wheel stops at a number that is BOTH a perfect square AND a prime number, the student is declared 'Champion' (special prize).

Based on the above situation, answer the following questions:
(i) What is the probability that the student wins a Gold Medal?
(ii) What is the probability that the student wins a Silver Medal?
(iii) What is the probability that the student wins a Bronze Medal?
        OR
(iii) A student claims: 'It is impossible to win the Champion prize in this game.' Is the student's claim correct? Justify your answer with mathematical reasoning and also find the probability of winning the Champion prize.

Accepted Answer

Total number of equally likely outcomes = 40

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(i) Probability of winning a Gold Medal [1 mark]
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Perfect squares from 1 to 40:
1, 4, 9, 16, 25, 36

⟹ Number of favourable outcomes = 6

∴ P(Gold Medal) = 6/40 = 3/20

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(ii) Probability of winning a Silver Medal [1 mark]
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Prime numbers greater than 20 and up to 40:
23, 29, 31, 37

⟹ Number of favourable outcomes = 4

∴ P(Silver Medal) = 4/40 = 1/10

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
(iii) Probability of winning a Bronze Medal [2 marks]
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A Bronze Medal requires a number that is a multiple of 6 but NOT a multiple of 4.

Multiples of 6 from 1 to 40:
6, 12, 18, 24, 30, 36

Now checking which of these are also multiples of 4:
⟹ 12 = 4 × 3 ✓ (multiple of 4, EXCLUDE)
⟹ 24 = 4 × 6 ✓ (multiple of 4, EXCLUDE)
⟹ 36 = 4 × 9 ✓ (multiple of 4, EXCLUDE)

Remaining multiples of 6 that are NOT multiples of 4:
6, 18, 30

⟹ Number of favourable outcomes = 3

∴ P(Bronze Medal) = 3/40

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OR
(iii) Champion Prize — Is the claim correct? [2 marks]
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The Champion prize requires a number that is BOTH a perfect square AND a prime number.

Perfect squares from 1 to 40: 1, 4, 9, 16, 25, 36

For a number to be both a perfect square and a prime, it must have exactly one prime factor (definition of prime) but a perfect square n = k² has k as a factor other than 1 and n itself whenever k > 1.
⟹ The only candidate would be 1, but 1 is neither prime nor composite.
⟹ No number from 1 to 40 is both a perfect square and a prime.

⟹ Number of favourable outcomes = 0

∴ P(Champion prize) = 0/40 = 0

Since P(Champion prize) = 0, this is an impossible event.

∴ The student's claim is CORRECT. It is indeed impossible to win the Champion prize in this game, because no natural number can be simultaneously a perfect square (composite or 1) and a prime number. Hence proved.

Question 6

A school library conducted a survey on the reading habits of its students. The librarian numbered all the books in the library from 1 to 120. She then randomly selects one book to recommend to a student.

Based on the above situation, answer the following questions:

(i) Find the probability that the number on the selected book is a perfect square.

(ii) Find the probability that the number on the selected book is divisible by both 4 and 6.

(iii) Find the probability that the number on the selected book has its unit digit as 5 OR the number is a multiple of 11.

OR

(iii) Find the probability that the number on the selected book is divisible by 7 but NOT divisible by 3.

Accepted Answer

(i) Total number of outcomes = 120

Perfect squares from 1 to 120: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121 — but 121 > 120, so perfect squares are:
1, 4, 9, 16, 25, 36, 49, 64, 81, 100

⟹ Number of favourable outcomes = 10

∴ P(perfect square) = 10/120 = 1/12

(ii) Total number of outcomes = 120

A number divisible by both 4 and 6 must be divisible by LCM(4, 6) = 12.

Multiples of 12 from 1 to 120: 12, 24, 36, 48, 60, 72, 84, 96, 108, 120

⟹ Number of favourable outcomes = 10

∴ P(divisible by both 4 and 6) = 10/120 = 1/12

(iii) Total number of outcomes = 120

Let A = event that unit digit is 5.
Numbers from 1 to 120 with unit digit 5: 5, 15, 25, 35, 45, 55, 65, 75, 85, 95, 105, 115
⟹ n(A) = 12

Let B = event that number is a multiple of 11.
Multiples of 11 from 1 to 120: 11, 22, 33, 44, 55, 66, 77, 88, 99, 110
⟹ n(B) = 10

A ∩ B = numbers with unit digit 5 AND divisible by 11 = {55}
⟹ n(A ∩ B) = 1

By the Addition Rule of Probability:
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

⟹ P(A ∪ B) = 12/120 + 10/120 − 1/120

⟹ P(A ∪ B) = 21/120 = 7/40

∴ P(unit digit is 5 OR multiple of 11) = 7/40

OR

(iii) Total number of outcomes = 120

Multiples of 7 from 1 to 120: 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98, 105, 112, 119
⟹ n(multiples of 7) = 17

Numbers divisible by both 7 and 3 must be divisible by LCM(7, 3) = 21.
Multiples of 21 from 1 to 120: 21, 42, 63, 84, 105
⟹ n(multiples of 21) = 5

Numbers divisible by 7 but NOT divisible by 3
⟹ Favourable outcomes = 17 − 5 = 12

∴ P(divisible by 7 but not by 3) = 12/120 = 1/10

Question 7

A school is organising a 'Science Talent Hunt' event. Students are given numbered tokens from 1 to 40 to participate in a lucky draw at the end. One token is drawn at random.

Based on the above situation, answer the following questions:
(i) Find the probability that the token drawn has a number which is a perfect square.
(ii) Find the probability that the token drawn has a number which is divisible by 6.
(iii) Find the probability that the token drawn has a number which is a prime number greater than 30.
     OR
     Find the probability that the token drawn has a number which is divisible by both 3 and 5.

Accepted Answer

Total number of tokens = 40
∴ Total number of equally likely outcomes = 40

(i) Perfect squares from 1 to 40: 1, 4, 9, 16, 25, 36
∴ Number of favourable outcomes = 6

P(token is a perfect square) = 6/40 = 3/20

∴ The probability that the token drawn is a perfect square = 3/20

(ii) Numbers from 1 to 40 divisible by 6: 6, 12, 18, 24, 30, 36
∴ Number of favourable outcomes = 6

P(token is divisible by 6) = 6/40 = 3/20

∴ The probability that the token drawn is divisible by 6 = 3/20

(iii) Prime numbers greater than 30 and up to 40: 31, 37
(Note: 32 = 2⁵, 33 = 3×11, 34 = 2×17, 35 = 5×7, 36 = 6², 38 = 2×19, 39 = 3×13, 40 = 2³×5 — none of these are prime)
∴ Number of favourable outcomes = 2

P(token is a prime number greater than 30) = 2/40 = 1/20

∴ The probability that the token drawn is a prime number greater than 30 = 1/20

OR

Numbers from 1 to 40 divisible by both 3 and 5 are numbers divisible by LCM(3, 5) = 15.
Numbers divisible by 15 from 1 to 40: 15, 30
∴ Number of favourable outcomes = 2

P(token is divisible by both 3 and 5) = 2/40 = 1/20

∴ The probability that the token drawn is divisible by both 3 and 5 = 1/20

Question 8

A school library has a box containing activity cards numbered 1 to 40. During a free period, a student picks one card at random. Read the following questions and answer them.

(i) What is the probability that the number on the card is a multiple of 5?

(ii) What is the probability that the number on the card is a prime number less than 15?

(iii) The librarian says: 'The probability of picking a card with a perfect square number is the same as picking a card with a number divisible by both 3 and 5.' Is the librarian correct? Justify your answer with complete calculation.

OR

(iii) A second box is introduced containing cards numbered 1 to 40 as well. One card is drawn from each box simultaneously. What is the probability that both cards show even numbers?

Accepted Answer

Total number of cards = 40.
∴ Total number of equally likely outcomes = 40.

**(i) Probability that the number is a multiple of 5:**

Multiples of 5 from 1 to 40: 5, 10, 15, 20, 25, 30, 35, 40.
∴ Number of favourable outcomes = 8.

P(multiple of 5) = 8/40 = **1/5**

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**(ii) Probability that the number is a prime number less than 15:**

Prime numbers less than 15 (from 1 to 40): 2, 3, 5, 7, 11, 13.
∴ Number of favourable outcomes = 6.

P(prime number less than 15) = 6/40 = **3/20**

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**(iii) [Main Option]**

**Perfect square numbers** from 1 to 40:
1, 4, 9, 16, 25, 36.
∴ Number of perfect squares = 6.

P(perfect square) = 6/40 = **3/20**

**Numbers divisible by both 3 and 5** (i.e., divisible by LCM(3, 5) = 15) from 1 to 40:
15, 30.
∴ Number of such cards = 2.

P(divisible by both 3 and 5) = 2/40 = **1/20**

Since 3/20 ≠ 1/20, the two probabilities are **not equal**.

∴ The librarian is **incorrect**. The probability of picking a perfect square (3/20) is not the same as picking a number divisible by both 3 and 5 (1/20).

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**(iii) [OR Option]**

For each box, total cards = 40.
Even numbers from 1 to 40: 2, 4, 6, …, 40.
∴ Number of even-numbered cards in each box = 20.

P(even card from Box 1) = 20/40 = 1/2.

P(even card from Box 2) = 20/40 = 1/2.

Since draws from the two boxes are independent events:

P(both cards show even numbers) = P(even from Box 1) × P(even from Box 2)

⟹ P = 1/2 × 1/2

∴ P(both cards even) = **1/4**

Question 9

Assertion (A): The probability of a sure event is 1.
Reason (R): The probability of any event lies between 0 and 1.

Accepted Answer

Option (b) is correct.

Explanation: Assertion (A) states that the probability of a sure event is 1, which is true, since a sure event always occurs and P(E) = Favourable outcomes / Total outcomes = Total outcomes / Total outcomes = 1. Reason (R) states that the probability of any event lies between 0 and 1 (inclusive), i.e. 0 ≤ P(E) ≤ 1, which is also true. However, R is not the correct explanation of A, because R merely gives the range of probability for any event and does not specifically explain why a sure event has probability exactly equal to 1; these are two independent standard results in probability. ∴ Both A and R are true but R is not the correct explanation of A.

Question 10

In a class survey, it is found that every eighth student owns a bicycle. The probability that a randomly selected student does NOT own a bicycle is:

Accepted Answer

Option (B) is correct.

Explanation: Using P(E) + P(E') = 1, since every eighth student owns a bicycle, P(owns bicycle) = 1/8. ∴ P(does NOT own bicycle) = 1 − 1/8 = 7/8.

Question 11

A bag contains 5 red balls, 3 blue balls, and 2 green balls. A ball is drawn at random from the bag. Find the probability that the ball drawn is NOT green.

Accepted Answer

Option (D) is correct.

Explanation: P(E) + P(E') = 1. Total number of balls = 5 + 3 + 2 = 10; number of green balls = 2 ⟹ P(green) = 2/10 = 1/5 ⟹ P(NOT green) = 1 − 1/5 = **4/5**.

Question 12

The probability that it will rain on a particular day is 0.83. What is the probability that it will NOT rain on that day?

Accepted Answer

Option (B) is correct.

**Explanation:** Using the complementary probability rule, P(E) + P(E') = 1. ⟹ P(not rain) = 1 − P(rain) = 1 − 0.83 = 0.17.

Question 13

A bag contains cards numbered 11 to 60. A card is drawn at random from the bag. The probability that the drawn card has a number which is a multiple of 7 is:

Accepted Answer

Option (C) is correct.

**Explanation:** P(E) = Favourable outcomes / Total outcomes. Cards numbered 11 to 60 give a total of 50 cards. Multiples of 7 in this range: 14, 21, 28, 35, 42, 49, 56 — 7 favourable outcomes. ∴ P(multiple of 7) = 7/50.

Question 14

Assertion (A): The probability of a sure event is 1.
Reason (R): The probability of any event lies between 0 and 1.

Accepted Answer

Option (b) is correct.

Explanation: Assertion (A) is true — by definition (axiom of probability), a sure event contains all possible outcomes, so P(sure event) = Favourable outcomes / Total outcomes = n/n = 1. Reason (R) is also true — for any event E, 0 ≤ P(E) ≤ 1. However, R merely states the general range of probability and does not explain why a sure event specifically has probability 1; that follows from the definition of a sure event, not from the range property. ∴ Both A and R are true but R is not the correct explanation of A.

Question 15

If G is an event such that P(G) = 0.64, then P(Ḡ) is equal to:

Accepted Answer

Option (A) is correct.

Explanation: By the complementary event rule, P(G̅) = 1 − P(G) ⟹ P(G̅) = 1 − 0.64 = 0.36.

Question 16

A leap year is selected at random. What is the probability that it contains 53 Mondays?

Accepted Answer

Option (A) is correct.

Explanation: A leap year has 366 days = 52 complete weeks + 2 extra days. The 2 extra days form one of 7 equally likely pairs: (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun), (Sun, Mon); total outcomes = 7. For 53 Mondays, the extra pair must contain a Monday: favourable pairs are (Mon, Tue) and (Sun, Mon), giving 2 favourable outcomes. ∴ P(53 Mondays) = 2/7.

Question 17

The digits 1, 2, 3, 4, 5, 6, 7, 8, 9 are written on separate slips and placed in a box. One slip is drawn at random. What is the probability that the slip does NOT bear an odd number?

Accepted Answer

Option (A) is correct.

**Explanation:** P(E) = Favourable outcomes / Total outcomes. Total slips = 9. Odd numbers (1, 3, 5, 7, 9) = 5, so slips NOT bearing an odd number (i.e., even: 2, 4, 6, 8) = 4. ∴ P(not odd) = 4/9.

Question 18

Assertion (A): The probability of a sure event is 1.
Reason (R): The probability of any event lies between 0 and 1.

Accepted Answer

Option (b) is correct.

Explanation: Assertion A is true — by definition, a sure (certain) event always occurs, so its probability = 1. Reason R is also true — for any event E, 0 ≤ P(E) ≤ 1. However, R is a general boundary property of probability and does not explain why a sure event specifically has probability 1; hence R is not the correct explanation of A.

Question 19

A digit is chosen at random from the digits 1 to 9. What is the probability that the chosen digit is NOT a perfect square?

Accepted Answer

Option (A) is correct.

Explanation: P(E) = Favourable outcomes / Total outcomes. Total digits from 1 to 9 = 9. Perfect squares among these digits are 1, 4, 9 — giving 3 favourable outcomes for "perfect square." ⟹ Digits that are NOT perfect squares = 9 − 3 = 6. ∴ P(not a perfect square) = 6/9 = **2/3**.

Question 20

A box contains 12 defective bulbs and some good bulbs. If the probability of selecting a good bulb at random from the box is 3/4, then the probability of selecting a defective bulb is:

Accepted Answer

Option (A) is correct.

**Explanation:** Using the complementary probability rule, P(E) + P(E') = 1. Since P(selecting a good bulb) = 3/4, P(selecting a defective bulb) = 1 − 3/4 = **1/4**.

Probability: Class 10 Maths Practice Questions

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