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MHT CET Probability & Combinatorics Questions

Practise probability & combinatorics (mathematics) for the Maharashtra Common Entrance Test: 10 multiple-choice questions, each with the correct answer and a worked explanation. Click an option to check it, or “New questions” for another set.

Probability & Combinatorics practice

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  1. Q1.In how many ways can a committee of 2 be chosen from 8 people?

    Show answer & explanation

    Answer: B — 28

    Order does not matter → combinations: 8C2 = 8! / (2!·6!) = 28.

  2. Q2.In how many ways can a committee of 3 be chosen from 7 people?

    Show answer & explanation

    Answer: D — 35

    Order does not matter → combinations: 7C3 = 7! / (3!·4!) = 35.

  3. Q3.Two fair dice are thrown. What is the probability that the sum is 8?

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    Answer: B — 5/36

    5 of the 36 equally likely outcomes give a sum of 8, so P = 5/36 = 5/36.

  4. Q4.Two fair dice are thrown. What is the probability that the sum is 10?

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    Answer: B — 1/12

    3 of the 36 equally likely outcomes give a sum of 10, so P = 3/36 = 1/12.

  5. Q5.Two fair dice are thrown. What is the probability that the sum is 5?

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    Answer: A — 1/9

    4 of the 36 equally likely outcomes give a sum of 5, so P = 4/36 = 1/9.

  6. Q6.In how many ways can the letters of the word SQUARE be arranged?

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    Answer: D — 720

    All 6 letters are different, so arrangements = 6! = 720.

  7. Q7.In how many ways can a committee of 3 be chosen from 9 people?

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    Answer: C — 84

    Order does not matter → combinations: 9C3 = 9! / (3!·6!) = 84.

  8. Q8.Two fair dice are thrown. What is the probability that the sum is 9?

    Show answer & explanation

    Answer: B — 1/9

    4 of the 36 equally likely outcomes give a sum of 9, so P = 4/36 = 1/9.

  9. Q9.In how many ways can a committee of 4 be chosen from 10 people?

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    Answer: B — 210

    Order does not matter → combinations: 10C4 = 10! / (4!·6!) = 210.

  10. Q10.3 fair coins are tossed. What is the probability of at least one head?

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    Answer: D — 7/8

    P(no head) = (1/2)^3 = 1/8; P(at least one head) = 1 − 1/8 = 7/8.

Free MHT CET Probability & Combinatorics Questions by pGento

How to approach probability & combinatorics questions

Probability = favourable ÷ total outcomes. List small sample spaces (two dice = 36 outcomes). Use nPr when order matters and nCr when it does not; divide by repeats for repeated letters.

About MHT CET

Maharashtra Common Entrance Test — conducted by State Common Entrance Test Cell, Maharashtra. State entrance for engineering, pharmacy and agriculture courses in Maharashtra.

Original practice questions written for this site in the style and difficulty of each exam’s syllabus. They are not past papers and the site is not affiliated with any exam body. Exam patterns, marking schemes and syllabi change — always check the latest official notification.

MHT CET practice topics

TopicSectionQuestion source
AlgebraMathematicsCalculated (new numbers every set)
Probability & CombinatoricsMathematicsCalculated (new numbers every set)
PhysicsScienceCalculated + question bank
ChemistryScienceCalculated + question bank
BiologyScienceQuestion bank

Frequently asked questions

In how many ways can a committee of 2 be chosen from 8 people?
Answer: B — 28. Order does not matter → combinations: 8C2 = 8! / (2!·6!) = 28.
In how many ways can a committee of 3 be chosen from 7 people?
Answer: D — 35. Order does not matter → combinations: 7C3 = 7! / (3!·4!) = 35.
Two fair dice are thrown. What is the probability that the sum is 8?
Answer: B — 5/36. 5 of the 36 equally likely outcomes give a sum of 8, so P = 5/36 = 5/36.
Are these real past exam papers?
No. Every question is an original practice question written in the style, syllabus and difficulty of the exam, with a worked explanation. The site is not affiliated with any exam body.
MHT CET Probability & Combinatorics Questions

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