In this chapter, we provide RD Sharma Class 10 Ex 13.2 Solutions Chapter 13 Probability for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 10 Ex 13.2 Solutions Chapter 13 Probability pdf, Now you will get step by step solution to each question.

RD Sharma Solutions for Class 10 Chapter 13 Probability Ex 13.2 Download PDF

Chapter 13: Probability Exercise – 13.2

Question: 1

Suppose you drop a tie at random on the rectangular region shown in fig. below. What is the probability that it will land inside the circle with diameter 1 m?

Solution:

Area of a circle with radius 0.5 m A circle = (0.5)² = 0.25 πm² 

Area of rectangle = 3 x 2 = 6m²

The probability that tie will land inside the circle with diameter 1m

Question: 2

In the accompanying diagram, a fair spinner is placed at the center O of the circle. Diameter AOB and radius OC divide the circle into three regions labeled X, Y and Z.? If ∠BOC = 45°. What is the probability that the spinner will land in the region X?

Solution:

Given, ∠BOC = 45° 

∠AOC = 180 – 45 = 135°

Area of circle = πr²

Area of region x = θ/360 × πr² 

= 135/360 × πr² 

= 3/8 × πr² 

The probability that the spinner will land in the region

Question: 3

A target is shown in fig. below consists of three concentric circles of radii, 3, 7 and 9 cm respectively. A dart is thrown and lands on the target. What is the probability that the dart will land on the shaded region?

Solution:

1^st circle – with radius 3

2^nd circle – with radius 7

3^rd circle – with radius 9

Area of 1^st circle = π(3)² = 9π 

Area of 2^nd circle = π(7)² = 49π 

Area of 3rd circle = π(9)² = 81π 

Area of shaded region = Area of 2^nd circle – Area of 1^st circle

= 49π − 9π 

= 40π 

Probability that it will land on the shaded region

Question: 4

In below fig. points A, B, C and D are the centers of four circles that each has a radius of length one unit. If a point is selected at random from the interior of square ABCD. What is the probability that the point will be chosen from the shaded region?

Solution:

Radius of circle = 1 cm

Length of side of square = 1 + 1 = 2 cm

Area of square = 2 × 2 = 4 cm² 

Area of shaded region = Area of a square – 4 × Area of the quadrant

Probability that the point will be chosen from the shaded region

Since geometrical probability, 

Question: 5

In the fig. below, JKLM is a square with sides of length 6 units. Points A and B are the midpoints of sides KL and LM respectively. If a point is selected at random from the interior of the square. What is the probability that the point will be chosen from the interior of triangle JAB?

Solution:

JKLM is a square with sides of length 6 units.

Points A and B are the midpoints of sides KL and ML, respectively.

If a point is selected at random from the interior of the square.

We have to find the probability that the point will be chosen from the interior of ΔJAB.

Now, Area of square JKLM is equal to 6² = 36 sq.units

Now, we have ar(ΔKAJ) = 1/2 × AK × KJ 

= 1/2 × 3 × 6 

= 9 unit² 

ar(ΔJMB) = 1/2 × JM × BM 

= 1/2 × 6 × 3 

= 9 unit² 

ar(ΔAJB) = 1/2 × AL × BL 

=1/2 × 3 × 3

= 9/2 unit² 

Now, an area of the triangle AJB ar(ΔAJB) = 3 × 9/2 

= 27/2 unit² 

We know that:

Hence, the probability that the point will be chosen from the interior of ΔAJB = 3/8.

Question: 6

In the fig. below, a square dartboard is shown. The length of a side of the larger square is 1.5 times the length of a side of the smaller square. If a dart is thrown and lands on the larger square. What is the probability that it will land in the interior of the smaller square?

Solution:

Let, the length of the side of smaller square = a

The a length of a side of bigger square = 1.5a

Area of smaller square = a² 

Area of bigger square = (1.5)²a² = 2.25a² 

Probability that dart will land in the interior of the smaller square

Geometrical probability,

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