Practice MCQs for Class 10 Mathematics IMO Full Syllabus
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These 15 IMO-style multiple choice questions on Probability for Class 10 test the basic definition of probability with coins, dice, playing cards, marbles, lettered and numbered cards, probability of an event and its complement, geometric probability and matching probabilities of events, with everyday situations such as leap years, lottery tickets, defective bulbs and shop visits. Each question has the correct answer with a step-by-step explanation and a short teacher's note, so you can practise for the International Mathematics Olympiad and check your working.
Mathematical Reasoning
(a) \( \dfrac{9}{100} \)
(b) \( \dfrac{1}{25} \)
(c) \( \dfrac{7}{100} \)
(d) \( \dfrac{3}{100} \)
Show Answer & Explanation
Answer: (d) \( \dfrac{3}{100} \)
Explanation:
1. A perfect square divisible by 9 must be the square of a multiple of 3.
2. Such numbers up to 100 are 3² = 9, 6² = 36 and 9² = 81, so there are 3 favourable outcomes.
3. Probability = \( \dfrac{3}{100} \).
Teacher's Note:
1. If 9 divides n² then 3 divides n, so list the squares of 3, 6, 9.
2. Check that 12² = 144 is already beyond 100.
(a) \( \dfrac{26}{49} \)
(b) \( \dfrac{23}{49} \)
(c) \( \dfrac{13}{26} \)
(d) \( \dfrac{23}{52} \)
Show Answer & Explanation
Answer: (b) \( \dfrac{23}{49} \)
Explanation:
1. Cards left = 52 − 3 = 49.
2. Black cards = 13 spades + 13 clubs = 26 originally; 3 spades are lost, so 26 − 3 = 23 black cards remain.
3. Probability = \( \dfrac{23}{49} \).
Teacher's Note:
1. Spades and clubs are both black, so there are 26 black cards in a full pack.
2. Remember to reduce both the favourable cases and the total.
(a) \( \dfrac{13}{49} \)
(b) \( \dfrac{10}{49} \)
(c) \( \dfrac{3}{49} \)
(d) \( \dfrac{1}{49} \)
Show Answer & Explanation
Answer: (b) \( \dfrac{10}{49} \)
Explanation:
1. Cards remaining = 52 − 3 = 49.
2. Clubs remaining = 13 − 3 = 10.
3. Probability of a club = \( \dfrac{10}{49} \).
Teacher's Note:
1. Removed clubs lower both the number of clubs and the total number of cards.
2. Do not keep 52 as the denominator.
(i) divisible by 5.
(ii) a number which is a perfect square.
(a) (i) \( \dfrac{5}{24} \), (ii) \( \dfrac{1}{24} \)
(b) (i) \( \dfrac{5}{24} \), (ii) \( \dfrac{1}{12} \)
(c) (i) \( \dfrac{1}{12} \), (ii) \( \dfrac{5}{12} \)
(d) None of these
Show Answer & Explanation
Answer: (b) (i) \( \dfrac{5}{24} \), (ii) \( \dfrac{1}{12} \)
Explanation:
1. Number of cards = 60 − 13 + 1 = 48.
2. (i) Multiples of 5: 15, 20, 25, 30, 35, 40, 45, 50, 55, 60 are 10 numbers, so probability = \( \dfrac{10}{48} \) = \( \dfrac{5}{24} \).
3. (ii) Perfect squares: 16, 25, 36, 49 are 4 numbers, so probability = \( \dfrac{4}{48} \) = \( \dfrac{1}{12} \).
Teacher's Note:
1. The count from 13 to 60 is 60 − 13 + 1 = 48, not 60 − 13 = 47 and not 60 − 12 + 1.
2. List the favourable numbers so none is missed.
(a) \( \dfrac{1}{4} \)
(b) \( \dfrac{1}{3} \)
(c) \( \dfrac{4}{9} \)
(d) \( \dfrac{7}{9} \)
Show Answer & Explanation
Answer: (d) \( \dfrac{7}{9} \)
Explanation:
1. Total marbles = 3 + 4 + 2 = 9.
2. Marbles that are not orange = 3 + 4 = 7.
3. Probability = \( \dfrac{7}{9} \).
Teacher's Note:
1. Equivalently, P(not orange) = 1 − P(orange) = 1 − 2/9.
2. Count the favourable marbles directly to avoid slips.
(a) \( \dfrac{2}{5} \)
(b) \( \dfrac{1}{5} \)
(c) \( \dfrac{4}{5} \)
(d) \( \dfrac{1}{10} \)
Show Answer & Explanation
Answer: (a) \( \dfrac{2}{5} \)
Explanation:
1. Total marbles = 6 + 4 = 10.
2. Green marbles = 4.
3. Probability = \( \dfrac{4}{10} \) = \( \dfrac{2}{5} \).
Teacher's Note:
1. Always reduce the fraction to its simplest form.
2. Probability = favourable outcomes ÷ total outcomes.
(a) \( \dfrac{3}{11} \)
(b) \( \dfrac{5}{11} \)
(c) \( \dfrac{6}{11} \)
(d) \( \dfrac{7}{11} \)
Show Answer & Explanation
Answer: (c) \( \dfrac{6}{11} \)
Explanation:
1. ASSOCIATION has 11 letters: A, S, S, O, C, I, A, T, I, O, N.
2. The vowels are A, O, I, A, I, O, i.e. 6 of them (repeated letters are counted each time).
3. Probability = \( \dfrac{6}{11} \).
Teacher's Note:
1. Count repeated letters separately, because each position in the word is a possible outcome.
2. Count the total letters first (11), then the vowels.
(a) 12
(b) 6
(c) 9
(d) 11
Show Answer & Explanation
Answer: (a) 12
Explanation:
1. Blue marbles = \( \dfrac{1}{3} \) × 54 = 18.
2. Green marbles = \( \dfrac{4}{9} \) × 54 = 24.
3. White marbles = 54 − 18 − 24 = 12.
Teacher's Note:
1. Number of marbles of a colour = probability × total.
2. Alternatively, P(white) = 1 − 1/3 − 4/9 = 2/9, and 2/9 of 54 is 12.
(a) \( \dfrac{2}{9} \)
(b) \( \dfrac{3}{7} \)
(c) \( \dfrac{4}{9} \)
(d) \( \dfrac{5}{9} \)
Show Answer & Explanation
Answer: (a) \( \dfrac{2}{9} \)
Explanation:
1. Total outcomes = 36.
2. Doublets: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) are 6 outcomes. Total 4: (1,3), (3,1), (2,2) are 3 outcomes.
3. (2,2) is in both lists, so favourable outcomes = 6 + 3 − 1 = 8, and probability = \( \dfrac{8}{36} \) = \( \dfrac{2}{9} \).
Teacher's Note:
1. Do not count the outcome (2, 2) twice.
2. For ‘A or B’ use n(A) + n(B) − n(A and B).
Everyday Mathematics
(a) \( \dfrac{1}{50} \)
(b) \( \dfrac{1}{125} \)
(c) \( \dfrac{3}{125} \)
(d) \( \dfrac{3}{50} \)
Show Answer & Explanation
Answer: (a) \( \dfrac{1}{50} \)
Explanation:
1. Total tickets = 250 and winning tickets = 5.
2. Probability = \( \dfrac{5}{250} \) = \( \dfrac{1}{50} \).
Teacher's Note:
1. Each ticket is equally likely to be the one Kunal holds.
2. Simplify by dividing both numbers by 5.
(a) 0.45
(b) 0.98
(c) 0.57
(d) 0.85
Show Answer & Explanation
Answer: (b) 0.98
Explanation:
1. Non-defective bulbs = 600 − 12 = 588.
2. Probability = \( \dfrac{588}{600} \) = \( \dfrac{49}{50} \) = 0.98.
Teacher's Note:
1. P(defective) = 12/600 = 0.02, so P(non-defective) = 1 − 0.02 = 0.98.
2. Answers in decimals should match the exact fraction.
(a) \( \dfrac{1}{7} \)
(b) \( \dfrac{6}{7} \)
(c) \( \dfrac{4}{7} \)
(d) \( \dfrac{5}{7} \)
Show Answer & Explanation
Answer: (c) \( \dfrac{4}{7} \)
Explanation:
1. He uses either the car or the bicycle, so the two probabilities add up to 1.
2. P(bicycle) = 1 − \( \dfrac{3}{7} \) = \( \dfrac{4}{7} \).
Teacher's Note:
1. The two events cover all possibilities and cannot happen together.
2. This is the complement rule.
Achievers Section (HOTS)
(a) \( \dfrac{5}{8} \)
(b) \( \dfrac{7}{8} \)
(c) \( \dfrac{3}{4} \)
(d) \( \dfrac{3}{8} \)
Show Answer & Explanation
Answer: (d) \( \dfrac{3}{8} \)
Explanation:
1. Area of square JKLM = 6² = 36 sq. units; KA = AL = LB = BM = 3.
2. The three corner triangles outside △JAB have areas: △JKA = \( \dfrac{1}{2} \times 6 \times 3 \) = 9, △ALB = \( \dfrac{1}{2} \times 3 \times 3 \) = \( \dfrac{9}{2} \), △JMB = 9.
3. Area of △JAB = 36 − \( \left(9 + 9 + \dfrac{9}{2}\right) \) = \( \dfrac{27}{2} \).
4. Probability = \( \dfrac{27/2}{36} \) = \( \dfrac{3}{8} \).
Teacher's Note:
1. Geometric probability is area of the favourable region ÷ area of the whole region.
2. Subtract the three right triangles at the corners instead of using a formula for △JAB.
| Column-I | Column-II |
|---|---|
| (P) Sum as prime number | (i) \( \dfrac{11}{36} \) |
| (Q) Multiple of 2 on one dice and multiple of 3 on other dice | (ii) \( \dfrac{1}{12} \) |
| (R) Total of at least 10 | (iii) \( \dfrac{5}{12} \) |
| (S) Doublet of even numbers | (iv) \( \dfrac{1}{6} \) |
(a) P → (iv), Q → (i), R → (iii), S → (ii)
(b) P → (iv), Q → (iii), R → (i), S → (ii)
(c) P → (iii), Q → (ii), R → (iv), S → (i)
(d) P → (iii), Q → (i), R → (iv), S → (ii)
Show Answer & Explanation
Answer: (d) P → (iii), Q → (i), R → (iv), S → (ii)
Explanation:
1. (P) Prime sums 2, 3, 5, 7, 11 occur in 1 + 2 + 4 + 6 + 2 = 15 outcomes: \( \dfrac{15}{36} \) = \( \dfrac{5}{12} \) → (iii).
2. (Q) First die even and second a multiple of 3: 3 × 2 = 6; first a multiple of 3 and second even: 6; both counted: (6,6). So 6 + 6 − 1 = 11 outcomes: \( \dfrac{11}{36} \) → (i).
3. (R) Total 10, 11 or 12: 3 + 2 + 1 = 6 outcomes: \( \dfrac{6}{36} \) = \( \dfrac{1}{6} \) → (iv).
4. (S) Even doublets (2,2), (4,4), (6,6): \( \dfrac{3}{36} \) = \( \dfrac{1}{12} \) → (ii).
Teacher's Note:
1. In part (Q) the outcome (6, 6) fits both descriptions, so count it only once.
2. Check the answer by looking at which Column-II values are left for the last item.
(i) In a single throw of a dice, the probability of getting a number greater than 2 is P .
(ii) A card is drawn from a deck of 52 cards. The probability of drawing a red card is Q and a face card is R .
(iii) A bag contains 2 blue and 3 green marbles, then the probability of drawing red marble S .
(a) P: \( \dfrac{1}{3} \), Q: \( \dfrac{11}{26} \), R: \( \dfrac{2}{11} \), S: 1
(b) P: \( \dfrac{1}{6} \), Q: \( \dfrac{1}{4} \), R: \( \dfrac{3}{11} \), S: 0
(c) P: \( \dfrac{2}{3} \), Q: \( \dfrac{1}{2} \), R: \( \dfrac{3}{13} \), S: 0
(d) P: \( \dfrac{2}{3} \), Q: \( \dfrac{1}{2} \), R: \( \dfrac{3}{13} \), S: \( \dfrac{3}{5} \)
Show Answer & Explanation
Answer: (c) P: \( \dfrac{2}{3} \), Q: \( \dfrac{1}{2} \), R: \( \dfrac{3}{13} \), S: 0
Explanation:
1. (i) Numbers greater than 2 are 3, 4, 5, 6, so P = \( \dfrac{4}{6} \) = \( \dfrac{2}{3} \).
2. (ii) A deck has 26 red cards, so Q = \( \dfrac{26}{52} \) = \( \dfrac{1}{2} \), and 12 face cards (J, Q, K in 4 suits), so R = \( \dfrac{12}{52} \) = \( \dfrac{3}{13} \).
3. (iii) The bag has no red marble, so S = 0 (an impossible event).
Teacher's Note:
1. Face cards are the jack, queen and king of each suit, 12 in all (aces are not face cards).
2. An event that cannot happen has probability 0.
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IMO Full Syllabus Objective Questions & Solutions for Class 10 Mathematics
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FAQs
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