CBSE Class 9 Mathematics Probability MCQs Set 10

Multiple Choice Questions (MCQs) for Class 9 Mathematics: Chapter 15 Probability

Access targeted multiple-choice questions for Chapter 15 Probability designed to align with the latest CBSE academic syllabus for Class 9 Mathematics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Practice Chapter 15 Probability MCQs for Class 9 Mathematics

View or download the dedicated Chapter 15 Probability MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Mathematics evaluations.

Multiple Choice Questions

Question 1: 17 cards numbered 1, 2, 3, ..., 17 are mixed in a box and one card is drawn at random. The probability that the number on it is not divisible by both 3 and 2 (that is, not divisible by 6) is
(a) \( \frac{2}{17} \)
(b) \( \frac{5}{17} \)
(c) \( \frac{14}{17} \)
(d) \( \frac{15}{17} \)
Show Answer & Explanation

Answer: (d) \( \frac{15}{17} \)

Explanation:
1. Numbers divisible by both 2 and 3 are multiples of 6: 6 and 12.
2. So 17 - 2 = 15 numbers are not divisible by 6.
3. Probability = 15/17.

Teacher's Note:
1. Divisible by both 2 and 3 means divisible by 6.

Question 2: A company surveyed 2370 families at random about monthly income and vehicles owned.
Monthly income (Rs.)0 vehicles1 vehicle2 vehiclesAbove 2
Less than 700010180250
7000-100000270272
10000-130001609291
13000-1600024092925
16000 or more15808288
If a family is chosen, the probability that it earns Rs. 10000-13000 per month and owns exactly 2 vehicles is
(a) \( \frac{203}{790} \)
(b) \( \frac{64}{237} \)
(c) \( \frac{29}{2370} \)
(d) \( \frac{9}{790} \)
Show Answer & Explanation

Answer: (c) \( \frac{29}{2370} \)

Explanation:
1. The table entry is 29 and the total number of families is 2370.
2. Probability = 29/2370.

Teacher's Note:
1. Empirical probability = (number of trials in which the event happens) / (total number of trials).
2. Always check that the frequencies add up to the total number of trials.

Question 3: A company surveyed 2370 families at random about monthly income and vehicles owned.
Monthly income (Rs.)0 vehicles1 vehicle2 vehiclesAbove 2
Less than 700010180250
7000-100000270272
10000-130001609291
13000-1600024092925
16000 or more15808288
If a family is chosen, the probability that it earns Rs. 16000 or more and owns exactly 1 vehicle is
(a) \( \frac{44}{1185} \)
(b) \( \frac{58}{237} \)
(c) \( \frac{751}{2370} \)
(d) \( \frac{409}{2370} \)
Show Answer & Explanation

Answer: (b) \( \frac{58}{237} \)

Explanation:
1. The table entry is 580.
2. Probability = 580/2370 = 58/237.

Teacher's Note:
1. Empirical probability = (number of trials in which the event happens) / (total number of trials).
2. Always check that the frequencies add up to the total number of trials.

Question 4: A company surveyed 2370 families at random about monthly income and vehicles owned.
Monthly income (Rs.)0 vehicles1 vehicle2 vehiclesAbove 2
Less than 700010180250
7000-100000270272
10000-130001609291
13000-1600024092925
16000 or more15808288
If a family is chosen, the probability that it earns less than Rs. 7000 and owns no vehicle is
(a) \( \frac{1}{237} \)
(b) \( \frac{43}{474} \)
(c) \( \frac{6}{79} \)
(d) \( \frac{7}{1185} \)
Show Answer & Explanation

Answer: (a) \( \frac{1}{237} \)

Explanation:
1. The table entry is 10.
2. Probability = 10/2370 = 1/237.

Teacher's Note:
1. Empirical probability = (number of trials in which the event happens) / (total number of trials).
2. Always check that the frequencies add up to the total number of trials.

Question 5: A company surveyed 2370 families at random about monthly income and vehicles owned.
Monthly income (Rs.)0 vehicles1 vehicle2 vehiclesAbove 2
Less than 700010180250
7000-100000270272
10000-130001609291
13000-1600024092925
16000 or more15808288
If a family is chosen, the probability that it earns Rs. 13000-16000 and owns more than 2 vehicles is
(a) \( \frac{31}{158} \)
(b) \( \frac{29}{2370} \)
(c) \( \frac{1}{1185} \)
(d) \( \frac{5}{474} \)
Show Answer & Explanation

Answer: (d) \( \frac{5}{474} \)

Explanation:
1. The table entry is 25.
2. Probability = 25/2370 = 5/474.

Teacher's Note:
1. Empirical probability = (number of trials in which the event happens) / (total number of trials).
2. Always check that the frequencies add up to the total number of trials.

Question 6: A company surveyed 2370 families at random about monthly income and vehicles owned.
Monthly income (Rs.)0 vehicles1 vehicle2 vehiclesAbove 2
Less than 700010180250
7000-100000270272
10000-130001609291
13000-1600024092925
16000 or more15808288
If a family is chosen, the probability that it owns not more than 1 vehicle is
(a) \( \frac{1024}{1185} \)
(b) \( \frac{7}{1185} \)
(c) \( \frac{1031}{1185} \)
(d) \( \frac{154}{1185} \)
Show Answer & Explanation

Answer: (c) \( \frac{1031}{1185} \)

Explanation:
1. Families with 0 vehicles: 10 + 0 + 1 + 2 + 1 = 14.
2. Families with 1 vehicle: 180 + 270 + 609 + 409 + 580 = 2048.
3. Not more than 1 vehicle: 14 + 2048 = 2062, so the probability is 2062/2370 = 1031/1185.

Teacher's Note:
1. Empirical probability = (number of trials in which the event happens) / (total number of trials).
2. Always check that the frequencies add up to the total number of trials.

Question 7: A company surveyed 2370 families at random about monthly income and vehicles owned.
Monthly income (Rs.)0 vehicles1 vehicle2 vehiclesAbove 2
Less than 700010180250
7000-100000270272
10000-130001609291
13000-1600024092925
16000 or more15808288
If a family is chosen, the probability that it owns at least one vehicle is
(a) \( \frac{7}{1185} \)
(b) \( \frac{1178}{1185} \)
(c) \( \frac{1031}{1185} \)
(d) \( \frac{1024}{1185} \)
Show Answer & Explanation

Answer: (b) \( \frac{1178}{1185} \)

Explanation:
1. Families with no vehicle = 14.
2. At least one vehicle: 2370 - 14 = 2356.
3. Probability = 2356/2370 = 1178/1185.

Teacher's Note:
1. Empirical probability = (number of trials in which the event happens) / (total number of trials).
2. Always check that the frequencies add up to the total number of trials.

Question 8: The blood groups of 30 students of a class are: A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O, A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O. A student is selected at random. The probability that the blood group of the student is A is
(a) \( \frac{3}{10} \)
(b) \( \frac{1}{5} \)
(c) \( \frac{1}{10} \)
(d) \( \frac{2}{5} \)
Show Answer & Explanation

Answer: (a) \( \frac{3}{10} \)

Explanation:
1. Count: A = 9, B = 6, AB = 3, O = 12 (total 30).
2. P(A) = 9/30 = 3/10.

Teacher's Note:
1. Empirical probability = (number of trials in which the event happens) / (total number of trials).
2. Always check that the frequencies add up to the total number of trials.

Question 9: The blood groups of 30 students of a class are: A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O, A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O. A student is selected at random. The probability that the blood group of the student is B is
(a) \( \frac{3}{10} \)
(b) \( \frac{1}{10} \)
(c) \( \frac{2}{5} \)
(d) \( \frac{1}{5} \)
Show Answer & Explanation

Answer: (d) \( \frac{1}{5} \)

Explanation:
1. B occurs 6 times, so P(B) = 6/30 = 1/5.

Teacher's Note:
1. Empirical probability = (number of trials in which the event happens) / (total number of trials).
2. Always check that the frequencies add up to the total number of trials.

Question 10: The blood groups of 30 students of a class are: A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O, A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O. A student is selected at random. The probability that the blood group of the student is AB is
(a) \( \frac{1}{5} \)
(b) \( \frac{3}{10} \)
(c) \( \frac{1}{10} \)
(d) \( \frac{2}{5} \)
Show Answer & Explanation

Answer: (c) \( \frac{1}{10} \)

Explanation:
1. AB occurs 3 times, so P(AB) = 3/30 = 1/10.

Teacher's Note:
1. Empirical probability = (number of trials in which the event happens) / (total number of trials).
2. Always check that the frequencies add up to the total number of trials.

Question 11: The blood groups of 30 students of a class are: A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O, A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O. A student is selected at random. The probability that the blood group of the student is O is
(a) \( \frac{3}{10} \)
(b) \( \frac{2}{5} \)
(c) \( \frac{1}{5} \)
(d) \( \frac{1}{10} \)
Show Answer & Explanation

Answer: (b) \( \frac{2}{5} \)

Explanation:
1. O occurs 12 times, so P(O) = 12/30 = 2/5.

Teacher's Note:
1. Empirical probability = (number of trials in which the event happens) / (total number of trials).
2. Always check that the frequencies add up to the total number of trials.

Question 12: Over 200 working days the number of defective parts produced by a machine was recorded.
Number of defective parts012345678910111213
Days50322218121210101086622
The probability that tomorrow's output has no defective part is
(a) \( \frac{1}{4} \)
(b) \( \frac{3}{4} \)
(c) \( \frac{1}{5} \)
(d) \( \frac{4}{25} \)
Show Answer & Explanation

Answer: (a) \( \frac{1}{4} \)

Explanation:
1. Days with no defective part = 50.
2. Probability = 50/200 = 1/4.

Teacher's Note:
1. Empirical probability = (number of trials in which the event happens) / (total number of trials).
2. Always check that the frequencies add up to the total number of trials.

Question 13: Over 200 working days the number of defective parts produced by a machine was recorded.
Number of defective parts012345678910111213
Days50322218121210101086622
The probability that tomorrow's output has at least one defective part is
(a) \( \frac{1}{4} \)
(b) \( \frac{73}{100} \)
(c) \( \frac{3}{5} \)
(d) \( \frac{3}{4} \)
Show Answer & Explanation

Answer: (d) \( \frac{3}{4} \)

Explanation:
1. Days with at least one defective part = 200 - 50 = 150.
2. Probability = 150/200 = 3/4.

Teacher's Note:
1. Empirical probability = (number of trials in which the event happens) / (total number of trials).
2. Always check that the frequencies add up to the total number of trials.

Question 14: Over 200 working days the number of defective parts produced by a machine was recorded.
Number of defective parts012345678910111213
Days50322218121210101086622
The probability that tomorrow's output has not more than 5 defective parts is
(a) \( \frac{27}{100} \)
(b) \( \frac{67}{100} \)
(c) \( \frac{73}{100} \)
(d) \( \frac{3}{4} \)
Show Answer & Explanation

Answer: (c) \( \frac{73}{100} \)

Explanation:
1. Days with 0 to 5 defective parts: 50 + 32 + 22 + 18 + 12 + 12 = 146.
2. Probability = 146/200 = 73/100 = 0.73.

Teacher's Note:
1. Empirical probability = (number of trials in which the event happens) / (total number of trials).
2. Always check that the frequencies add up to the total number of trials.

Question 15: Over 200 working days the number of defective parts produced by a machine was recorded.
Number of defective parts012345678910111213
Days50322218121210101086622
The probability that tomorrow's output has more than 13 defective parts is
(a) \( \frac{1}{200} \)
(b) \( 0 \)
(c) \( \frac{1}{100} \)
(d) \( \frac{1}{4} \)
Show Answer & Explanation

Answer: (b) \( 0 \)

Explanation:
1. No day had more than 13 defective parts in the record.
2. The probability is 0.

Teacher's Note:
1. Empirical probability = (number of trials in which the event happens) / (total number of trials).
2. Always check that the frequencies add up to the total number of trials.

Question 16: Three coins were tossed together 100 times with the following results.
OutcomeNo headOne headTwo headsThree heads
Frequency2040337
The probability of getting no head is
(a) \( \frac{1}{5} \)
(b) \( \frac{2}{5} \)
(c) \( \frac{33}{100} \)
(d) \( \frac{7}{100} \)
Show Answer & Explanation

Answer: (a) \( \frac{1}{5} \)

Explanation:
1. No head occurred 20 times out of 100.
2. Probability = 20/100 = 1/5.

Teacher's Note:
1. Empirical probability = (number of trials in which the event happens) / (total number of trials).
2. Always check that the frequencies add up to the total number of trials.

Question 17: Three coins were tossed together 100 times with the following results.
OutcomeNo headOne headTwo headsThree heads
Frequency2040337
The probability of getting two heads is
(a) \( \frac{2}{5} \)
(b) \( \frac{7}{100} \)
(c) \( \frac{1}{5} \)
(d) \( \frac{33}{100} \)
Show Answer & Explanation

Answer: (d) \( \frac{33}{100} \)

Explanation:
1. Two heads occurred 33 times out of 100.
2. Probability = 33/100.

Teacher's Note:
1. Empirical probability = (number of trials in which the event happens) / (total number of trials).
2. Always check that the frequencies add up to the total number of trials.

Question 18: A bag contains 6 black, 7 red and 2 white balls. A ball is drawn at random. The probability that it is red is
(a) \( \frac{2}{5} \)
(b) \( \frac{2}{15} \)
(c) \( \frac{7}{15} \)
(d) \( \frac{8}{15} \)
Show Answer & Explanation

Answer: (c) \( \frac{7}{15} \)

Explanation:
1. Total balls = 15 and red balls = 7.
2. Probability = 7/15.

Teacher's Note:
1. Count all balls first.
2. Probability = favourable / total.

Question 19: A bag contains 6 black, 7 red and 2 white balls. A ball is drawn at random. The probability that it is black or white is
(a) \( \frac{7}{15} \)
(b) \( \frac{8}{15} \)
(c) \( \frac{2}{5} \)
(d) \( \frac{3}{5} \)
Show Answer & Explanation

Answer: (b) \( \frac{8}{15} \)

Explanation:
1. Black or white balls = 6 + 2 = 8.
2. Probability = 8/15.

Teacher's Note:
1. Add the favourable cases of the two colours.

Question 20: A bag contains 6 black, 7 red and 2 white balls. A ball is drawn at random. The probability that it is not black is
(a) \( \frac{3}{5} \)
(b) \( \frac{2}{5} \)
(c) \( \frac{7}{15} \)
(d) \( \frac{1}{3} \)
Show Answer & Explanation

Answer: (a) \( \frac{3}{5} \)

Explanation:
1. Balls that are not black = 7 + 2 = 9.
2. Probability = 9/15 = 3/5.

Teacher's Note:
1. P(not black) = 1 - 6/15.

Question 21: A die was thrown 400 times with the following results.
Outcome123456
Frequency756065706862
The probability of getting an odd number is
(a) \( \frac{12}{25} \)
(b) \( \frac{19}{40} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{13}{25} \)
Show Answer & Explanation

Answer: (d) \( \frac{13}{25} \)

Explanation:
1. Odd outcomes 1, 3, 5: 75 + 65 + 68 = 208.
2. Probability = 208/400 = 13/25.

Teacher's Note:
1. Add the frequencies of 1, 3 and 5.

Question 22: A die was thrown 400 times with the following results.
Outcome123456
Frequency756065706862
The probability of getting a prime number is
(a) \( \frac{203}{400} \)
(b) \( \frac{133}{400} \)
(c) \( \frac{193}{400} \)
(d) \( \frac{1}{2} \)
Show Answer & Explanation

Answer: (c) \( \frac{193}{400} \)

Explanation:
1. Prime outcomes 2, 3, 5: 60 + 65 + 68 = 193.
2. Probability = 193/400.

Teacher's Note:
1. 1 is not prime; 4 and 6 are not prime.

Question 23: A coin is tossed 1000 times. If the probability of getting a tail is \( \frac38 \), the number of times a head is obtained is
(a) 375
(b) 625
(c) 500
(d) 250
Show Answer & Explanation

Answer: (b) 625

Explanation:
1. P(head) = 1 - 3/8 = 5/8.
2. Number of heads = 5/8 of 1000 = 625.

Teacher's Note:
1. Use P(head) + P(tail) = 1.
2. Number of heads = probability times number of trials.

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Multiple Choice Questions (MCQs) for Class 9 Mathematics Chapter 15 Probability

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FAQs

Where can I access latest CBSE Class 9 Mathematics Probability MCQs Set 10?

You can get most exhaustive CBSE Class 9 Mathematics Probability MCQs Set 10 for free on StudiesToday.com. These MCQs for Class 9 Mathematics are updated for the 2026-27 academic session as per CBSE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Mathematics Class 9 material?

Yes, our CBSE Class 9 Mathematics Probability MCQs Set 10 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the CBSE paper is now competency-based.

How do practicing Mathematics MCQs help in scoring full marks in Class 9 exams?

By solving our CBSE Class 9 Mathematics Probability MCQs Set 10, Class 9 students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Mathematics.

Do you provide answers and explanations for CBSE Class 9 Mathematics Probability MCQs Set 10?

Yes, Mathematics MCQs for Class 9 have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused CBSE exams.

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