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Spinning Arrow and Geometric Probability | Class 10 Maths

Solve the NCERT spinning arrow probability problem and learn how to find probability using areas, with a die dropped on a rectangle containing a circle, for Class 10 Maths.

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Updated 25 October 2026

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Quick answer: For an arrow that is equally likely to stop at any of the numbers 11 to 88: P(8)=18P(8) = \frac{1}{8}, P(odd number)=48=12P(\text{odd number}) = \frac{4}{8} = \frac{1}{2}, P(number greater than 2)=68=34P(\text{number greater than } 2) = \frac{6}{8} = \frac{3}{4}, and P(number less than 9)=88=1P(\text{number less than } 9) = \frac{8}{8} = 1 (a sure event). When an object lands at random on a region, probability can also be found using areas: P=favourable areatotal areaP = \frac{\text{favourable area}}{\text{total area}}; for a die dropped on a 3 m×2 m3\text{ m} \times 2\text{ m} rectangle containing a circle of diameter 1 m1\text{ m}, P=π24≈0.13P = \frac{\pi}{24} \approx 0.13.

Most probability questions count outcomes: the faces of a die, the cards in a pack, or the numbers on a spinner. This guide solves the NCERT spinning arrow problem, and then looks at geometric probability, where probability is found by comparing areas instead of counting.

In CBSE Class 10 Mathematics, Chapter 14 (Probability), the spinning arrow problem is a well-known NCERT question.

Syllabus note: The spinning arrow problem is part of the NCERT exercise. The geometric (area) probability problem was an optional NCERT exercise and is not part of the current rationalised textbook. It is included here as enrichment.


What You Will Learn

  • How to find probabilities for a spinner with equally likely numbers
  • What a sure event is
  • How to find probability using areas
  • A solved example: a die dropped on a rectangle containing a circle

1. Problem 1: The Spinning Arrow (NCERT)

Question

A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1,2,3,4,5,6,7,81, 2, 3, 4, 5, 6, 7, 8, and these are equally likely outcomes. What is the probability that it will point at:

(i) 88?

(ii) an odd number?

(iii) a number greater than 22?

(iv) a number less than 99?

Solution

There are 88 equally likely outcomes: 1,2,3,4,5,6,7,81, 2, 3, 4, 5, 6, 7, 8.

(i) Pointing at 8. Only 11 favourable outcome:

P(8)=18P(8) = \frac{1}{8}

(ii) An odd number. The odd numbers are 1,3,5,71, 3, 5, 7, which is 44 outcomes:

P(odd)=48=12P(\text{odd}) = \frac{4}{8} = \frac{1}{2}

(iii) A number greater than 2. These are 3,4,5,6,7,83, 4, 5, 6, 7, 8, which is 66 outcomes:

P(greater than 2)=68=34P(\text{greater than } 2) = \frac{6}{8} = \frac{3}{4}

(iv) A number less than 9. All 88 numbers are less than 99:

P(less than 9)=88=1P(\text{less than } 9) = \frac{8}{8} = 1

Key point: An event that is certain to happen, such as getting a number less than 99 here, is called a sure event. Its probability is 11.


2. Geometric Probability: Using Areas (Enrichment)

When an object lands at random on a region, every point of the region is equally likely. Then:

P(landing in a part of the region)=area of that parttotal area of the regionP(\text{landing in a part of the region}) = \frac{\text{area of that part}}{\text{total area of the region}}


3. Problem 2: A Die Dropped on a Rectangle

Question

A die is dropped at random on a rectangular region measuring 3 m×2 m3\text{ m} \times 2\text{ m}. Inside it is a circle of diameter 1 m1\text{ m}. What is the probability that the die lands inside the circle?

Solution

Step 1: Total area.

Area of rectangle=3×2=6 m2\text{Area of rectangle} = 3 \times 2 = 6\text{ m}^2

Step 2: Area of the circle. The diameter is 1 m1\text{ m}, so the radius is 12 m\frac{1}{2}\text{ m}:

Area of circle=π(12)2=π4 m2\text{Area of circle} = \pi \left(\frac{1}{2}\right)^2 = \frac{\pi}{4}\text{ m}^2

Step 3: Probability.

P=π/46=π24≈0.13P = \frac{\pi/4}{6} = \frac{\pi}{24} \approx 0.13

Answer: The probability is π24\frac{\pi}{24}, about 0.130.13.


4. Summary Table

EventFavourableTotalProbability
Spinner points at 81818\frac{1}{8}
Odd number4812\frac{1}{2}
Number greater than 26834\frac{3}{4}
Number less than 98811 (sure event)
Die lands inside the circleArea π4\frac{\pi}{4}Area 66π24\frac{\pi}{24}

5. Exam Tips and Common Mistakes

Exam Tip: List the favourable outcomes before writing each probability, e.g. "odd numbers: 1, 3, 5, 7".

Exam Tip: It is fine to leave an answer in terms of π\pi, such as π24\frac{\pi}{24}, unless a decimal is asked for.

Common Mistake: Using the diameter as the radius. For a circle of diameter 1 m1\text{ m}, r=0.5 mr = 0.5\text{ m}. Using r=1 mr = 1\text{ m} gives an area four times too large.

Common Mistake: Counting 22 as "greater than 22". It is not; the numbers greater than 22 are 33 to 88.


Frequently Asked Questions

What is the probability that the spinning arrow points at an odd number?

There are 4 odd numbers out of 8, so the probability is 1/2.

What is the probability that the arrow points at a number less than 9?

All 8 numbers are less than 9, so the probability is 1. This is a sure event.

What is a sure event?

An event that is certain to happen. Its probability is 1.

How do you find probability using areas?

Divide the area of the favourable region by the total area of the region where the object can land.

What is the probability that a die dropped on a 3 m × 2 m rectangle lands in a circle of diameter 1 m?

π/24, which is about 0.13.

Concept Check

MEDIUM

A fraction becomes 13\frac{1}{3} when 1 is subtracted from the numerator, and it becomes 14\frac{1}{4} when 8 is added to its denominator. Find the fraction.

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