Class 10 Maths Chapter 14 Probability Extra Questions (HOTS)

Genuinely harder, HOTS-level practice for Class 10 Maths Chapter 14 (Probability), going beyond the textbook exercise with fresh scenarios and reverse-engineering problems. These Class 10 Mathematics Chapter 14 important questions are handy for last-minute exam practice.

Last Updated: September 23, 2026

How to Approach the HOTS Questions in This Chapter

1. Q1 (Assertion-Reason). Assertion (A): P(sum=13 on two dice)=0. Reason (R): P(impossible event)=0 always.
   Solution: Max sum on two dice is 6+6=12, so sum=13 is impossible, confirming A. R correctly states the general rule and explains A. Answer: Both A and R true, R explains A.
2. Q2 (Reverse-engineering). A bag has some red marbles and 18 green marbles. P(red)=2/5. Find the number of red marbles and total marbles.
   Solution: r/(r+18)=2/5 ⇒ 5r=2r+36 ⇒ r=12. Total=30 (12 red, 18 green). Check: 12/30=2/5 ✓
3. Q3 (Fresh scenario). Two dice thrown together. Find P(product of the two numbers is a perfect square).
   Solution: Total=36. Products that are perfect squares (1,4,9,16,25,36): (1,1)=1 way; (1,4)(4,1)(2,2)=3 ways for 4; (3,3)=1 for 9; (4,4)=1 for 16; (5,5)=1 for 25; (6,6)=1 for 36. Total favourable=8. P=8/36=2/9.
4. Q4 (Fresh scenario). A card drawn from 52. Find P(neither a king nor a queen).
   Solution: Exclude 4 kings+4 queens=8. Favourable=52-8=44. P=44/52=11/13.
5. Q5 (Fresh scenario). Four fair coins tossed together. Find P(exactly 2 heads), P(at least 3 heads), P(no head).
   Solution: Total=2⁴=16. Exactly 2 heads=⁴C₂=6 ⇒ 3/8. At least 3=4(exactly 3)+1(exactly4)=5 ⇒ 5/16. No head=1 (TTTT) ⇒ 1/16.
6. Q6 (Reverse-engineering with follow-up). A box has some red pens and 15 blue pens. P(red)=3/8. Find red pens and total. If 5 more blue pens are added, find the new P(red).
   Solution: r/(r+15)=3/8 ⇒ 8r=3r+45 ⇒ r=9. Total=24 (9 red, 15 blue). After adding 5 blue: total=29, red unchanged=9. New P(red)=9/29.
7. Q7 (Fresh scenario). Two dice thrown together. Find P(sum is a multiple of 4).
   Solution: Multiples of 4 possible: 4,8,12. Sum4=3 ways, Sum8=5 ways, Sum12=1 way. Total favourable=9. P=9/36=1/4.
8. Q8 (Fresh scenario). A letter is chosen at random from the word MATHEMATICS. Find P(vowel), P(letter is M), P(consonant).
   Solution: 11 letters: M=2,A=2,T=2,H=1,E=1,I=1,C=1,S=1. Vowels(A,E,I)=2+1+1=4 ⇒ P(vowel)=4/11. P(M)=2/11. Consonants=11-4=7 ⇒ P(consonant)=7/11.

More on This Chapter

• NCERT Solutions – Chapter 14
• Extra Questions (HOTS) – Chapter 14
• Revision Notes – Chapter 14
• Class 10 Maths Book (Catalog Page)

Frequently Asked Questions

What is the difference between a sample space and a favourable outcome in a probability problem?
A sample space is the complete set of all possible outcomes of an experiment, while a favourable outcome is one or more specific outcomes within that sample space that satisfy the condition being asked about.

Why does the probability of an event and the probability of its complement always add up to exactly one?
The event and its complement together cover every possible outcome in the sample space with no overlap, so their combined probability must account for the entire sample space, which has a total probability of exactly one.

Written by Satish

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